Thermal Properties of Matter — Competition Zone

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75° C. T is given by : (Given : room temperature = 30°C, specific heat of copper = 0.1 cal/gm°C)

1. 825°C
2. 800°C
3. 885°C
4. 1250°C

Answer

885°C

Reason — Given,

• Mass of the copper ball, m₁ = 100 g at temperature T
• Mass of the copper calorimeter, m₂ = 100 g at 30°C
• Mass of water, m₃ = 170 g at 30°C
• Final temperature of the system = 75°C
• Specific heat of copper, c_Cu = 0.1 cal g^-1 °C^-1; specific heat of water, c_w = 1 cal g^-1 °C^-1

The heat lost by the copper ball in cooling from T to 75°C is

$\text Q_1 = \text m_1\text c_{Cu}(\text T - 75) = 100 \times 0.1 \times (\text T - 75) \\[1em] = 10(\text T - 75)$

The heat gained by the calorimeter and the water in warming from 30°C to 75°C is

$\text Q_2 = (\text m_2\text c_{Cu} + \text m_3\text c_\text w)(75 - 30)$

$= \left[(100 \times 0.1) + (170 \times 1)\right] \times 45 \\[1em] = (10 + 170) \times 45 = 8100\ \text{cal}$

By the principle of calorimetry, heat lost = heat gained,

$10(\text T - 75) = 8100$

$\text T - 75 = 810 \quad \Rightarrow \quad \text T = 885^\circ \text C$

A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 10⁵ Nm^-2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is :

1. 104.3 J
2. 84.5 J
3. 42.2 J
4. 208.7 J

Answer

208.7 J

Reason — Given,

• Mass of water, m = 0.1 g at 100°C
• Heat supplied, Q = 54 cal
• Pressure, P = 1.013 × 10⁵ N m^-2
• Volume of steam produced, V₂ = 167.1 cc; volume of water, V₁ = 0.1 cc

Converting the heat supplied into joule, using 1 cal = 4.18 J,

$\text Q = 54 \times 4.18 = 225.7\ \text J$

The change in volume during vaporisation is

$\Delta \text V = \text V_2 - \text V_1 = 167.1 - 0.1 = 167\ \text{cc} \\[1em] = 167 \times 10^{-6}\ \text m^3$

The work done by the sample in expanding against the constant external pressure is

$\text W = \text P\Delta \text V = (1.013 \times 10^5) \times (167 \times 10^{-6}) \\[1em] = 16.92\ \text J$

By the first law of thermodynamics, the change in internal energy is

$\Delta \text U = \text Q - \text W = 225.7 - 16.92 \\[1em] = 208.8 \approx 208.7\ \text J$

When m₁ gram of ice at −10°C (specific heat = 0.5 cal g^-1 °C^-1) is added to m₂ gram of water at 50°C, finally no ice is left and the water is at 0°C. The value of latent heat of ice in cal g^-1, is :

1. $\dfrac{50\text m_2}{\text m_1} - 5$
2. $\dfrac{50\text m_1}{\text m_2} - 50$
3. $\dfrac{50\text m_2}{\text m_1}$
4. $\dfrac{5\text m_2}{\text m_1}$

Answer

$\dfrac{50\text m_2}{\text m_1} - 5$

Reason — Given,

• Mass of ice, m₁ gram at − 10°C, specific heat of ice = 0.5 cal g^-1 °C^-1
• Mass of water, m₂ gram at 50°C, specific heat of water = 1 cal g^-1 °C^-1
• Final temperature = 0°C, and no ice is left

Let L be the latent heat of ice.

The heat gained by the ice is used up first in warming it from − 10°C to 0°C and then in melting the whole of it at 0°C,

$\text Q_1 = \text m_1 \times 0.5 \times 10 + \text m_1\text L = 5\text m_1 + \text m_1\text L$

The heat lost by the water in cooling from 50°C to 0°C is

$\text Q_2 = \text m_2 \times 1 \times 50 = 50\text m_2$

By the principle of calorimetry, heat gained = heat lost,

$5\text m_1 + \text m_1\text L = 50\text m_2$

$\text m_1\text L = 50\text m_2 - 5\text m_1$

$\text L = \dfrac{50\text m_2}{\text m_1} - 5$

A thermally insulated vessel contains 150 g of water at 0°C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be closest to (Latent heat of vaporisation of water = 2.10 × 10⁶ J/kg, latent heat of fusion of water = 3.36 × 10⁵ J/kg) :

1. 150 g
2. 20 g
3. 130 g
4. 35 g

Answer

20 g

Reason — Given,

• Total mass of water, M = 150 g at 0°C
• Latent heat of vaporisation, L_v = 2.10 × 10⁶ J kg^-1 = 2100 J g^-1
• Latent heat of fusion, L_f = 3.36 × 10⁵ J kg^-1 = 336 J g^-1

Let m gram of water evaporate. Then the remaining (150 − m) gram turns into ice.

The vessel is thermally insulated, so no heat comes in from outside. The heat needed for the evaporation must therefore be supplied by the heat given out when the rest of the water freezes,

$\text m\text L_\text v = (150 - \text m)\text L_\text f$

Substituting the values,

$2100\text m = 336(150 - \text m)$

$2100\text m = 50400 - 336\text m$

$2436\text m = 50400$

$\text m = \dfrac{50400}{2436} = 20.7 \approx 20\ \text g$

When 100 g of a liquid A at 100°C is added to 50 g of a liquid B at temperature 75°C, the temperature of the mixture becomes 90°C. The temperature of the mixture, if 100 g of liquid A at 100°C is added to 50 g of liquid B at 50°C will be :

1. 60°C
2. 80°C
3. 70°C
4. 85°C

Answer

80°C

Reason — Given,

• First case : 100 g of A at 100°C mixed with 50 g of B at 75°C gives a mixture at 90°C
• Second case : 100 g of A at 100°C mixed with 50 g of B at 50°C

Let c_A and c_B be the specific heats of the two liquids.

First case : By the principle of calorimetry, heat lost by A = heat gained by B,

$100\text c_\text A(100 - 90) = 50\text c_\text B(90 - 75)$

$1000\text c_\text A = 750\text c_\text B \quad \Rightarrow \quad \text c_\text A = 0.75\text c_\text B$

Second case : Let T be the final temperature of the mixture,

$100\text c_\text A(100 - \text T) = 50\text c_\text B(\text T - 50)$

Substituting c_A = 0.75 c_B and cancelling c_B,

$100 \times 0.75 \times (100 - \text T) = 50(\text T - 50)$

$75(100 - \text T) = 50(\text T - 50)$

$7500 - 75\text T = 50\text T - 2500$

$125\text T = 10000 \quad \Rightarrow \quad \text T = 80^\circ \text C$

A metal ball of mass 0.1 kg is heated upto 500°C and dropped into a vessel of heat capacity 800 J/K and containing 0.5 kg water. The initial temperature of water and vessel is 30°C. What is the approximate percentage increment in the temperature of the water?

[Take specific heat capacities of water and metal as 4200 J kg^-1 K^-1 and 400 J kg^-1 K^-1]

1. 25%
2. 15%
3. 30%
4. 20%

Answer

20%

Reason — Given,

• Mass of the metal ball, m = 0.1 kg at 500°C, specific heat c = 400 J kg^-1 K^-1
• Heat capacity of the vessel = 800 J K^-1
• Mass of water, m_w = 0.5 kg at 30°C, specific heat c_w = 4200 J kg^-1 K^-1
• Initial temperature of the water and the vessel = 30°C

Let T be the final temperature of the mixture.

The heat lost by the metal ball is

$\text Q_1 = \text{mc}(500 - \text T) = 0.1 \times 400 \times (500 - \text T) \\[1em] = 40(500 - \text T)$

The heat gained by the water and the vessel is

$\text Q_2 = (\text m_\text w\text c_\text w + 800)(\text T - 30)$

$= \left[(0.5 \times 4200) + 800\right](\text T - 30) \\[1em] = 2900(\text T - 30)$

By the principle of calorimetry, heat lost = heat gained,

$40(500 - \text T) = 2900(\text T - 30)$

$20000 - 40\text T = 2900\text T - 87000$

$2940\text T = 107000 \quad \Rightarrow \quad \text T = 36.39^\circ \text C$

The rise in the temperature of the water is 36.39 − 30 = 6.39°C. Hence the percentage increment is

$\dfrac{6.39}{30} \times 100 = 21.3$

Ice at −20°C is added to 50 g of water at 40°C. When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to :

(Specific heat of water = 4.2 J/g°C, specific heat of ice = 2.1 J/g°C, latent heat of fusion of water at 0°C = 334 J/g)

1. 40 g
2. 50 g
3. 60 g
4. 100 g

Answer

40 g

Reason — Given,

• Mass of water, m_w = 50 g at 40°C, specific heat c_w = 4.2 J g^-1 °C^-1
• Ice added at − 20°C, specific heat of ice c_i = 2.1 J g^-1 °C^-1
• Latent heat of fusion, L = 334 J g^-1
• Final temperature = 0°C, with 20 g of ice still unmelted

Let m be the mass of ice added.

The heat released by the water in cooling from 40°C to 0°C is

$\text Q_1 = \text m_\text w\text c_\text w \times 40 = 50 \times 4.2 \times 40 = 8400\ \text J$

The heat absorbed by the ice is used up in two parts — first the whole mass m is warmed from − 20°C to 0°C, and then only (m − 20) gram of it melts, since 20 g remains unmelted,

$\text Q_2 = \text m\text c_\text i \times 20 + (\text m - 20)\text L$

$= 42\text m + 334(\text m - 20)$

By the principle of calorimetry, heat released = heat absorbed,

$42\text m + 334(\text m - 20) = 8400$

$42\text m + 334\text m - 6680 = 8400$

$376\text m = 15080 \quad \Rightarrow \quad \text m = 40.1 \approx 40\ \text g$

A uniform cylindrical rod of length L and radius r, is made from a material whose Young's modulus of elasticity is Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The co-efficient of volume expansion of the material of the rod is (nearly) equal to :

1. $\dfrac{9\text F}{\pi \text r^2 \text{YT}}$

2. $\dfrac{6\text F}{\pi \text r^2 \text{YT}}$

3. $\dfrac{3\text F}{\pi \text r^2 \text{YT}}$

4. $\dfrac{\text F}{3\pi \text r^2 \text{YT}}$

Answer

$\dfrac{3\text F}{\pi \text r^2 \text{YT}}$

Reason — Given,

• Length of the rod = L, radius = r, so area of cross-section A = πr²
• Young's modulus = Y, rise in temperature = T
• Compressional force = F, and the length of the rod remains unchanged

On heating, the rod tends to expand in length by

$\Delta \text L_1 = \alpha\text L\text T$

where α is the coefficient of linear expansion. At the same time the compressional force tends to shorten it. From the definition of Young's modulus,

$\text Y = \dfrac{\text F/\text A}{\Delta \text L_2/\text L} \quad \Rightarrow \quad \Delta \text L_2 = \dfrac{\text{FL}}{\text{AY}}$

Since the length of the rod remains unchanged, the expansion due to heating is exactly balanced by the compression due to the force,

$\alpha\text L\text T = \dfrac{\text{FL}}{\text{AY}}$

$\alpha = \dfrac{\text F}{\text{AYT}} = \dfrac{\text F}{\pi \text r^2 \text{YT}}$

The coefficient of volume expansion is γ = 3α, so

$\gamma = \dfrac{3\text F}{\pi \text r^2 \text{YT}}$

Two rods A and B of identical dimensions are at temperature 30°C. If A is heated up to 180°C and B up to T°C, then new lengths are the same. If the ratio of the co-efficient of linear expansion of A and B is 4 : 3 then the value of T is :

1. 230°C
2. 270°C
3. 200°C
4. 250°C

Answer

230°C

Reason — Given,

• Both rods are of identical dimensions and are initially at 30°C
• Rod A is heated to 180°C, so ΔT_A = 180 − 30 = 150°C
• Rod B is heated to T°C, so ΔT_B = T − 30
• Ratio of the coefficients of linear expansion, α_A : α_B = 4 : 3

Since the rods are of identical dimensions, they have the same original length L. Their new lengths will be the same only if the increases in their lengths are equal,

$\text L\alpha_\text A\Delta \text T_\text A = \text L\alpha_\text B\Delta \text T_\text B$

$\alpha_\text A \times 150 = \alpha_\text B(\text T - 30)$

$\text T - 30 = \dfrac{\alpha_\text A}{\alpha_\text B} \times 150 = \dfrac{4}{3} \times 150 = 200$

$\text T = 200 + 30 = 230^\circ \text C$

A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is :

(α_Cu = 1.7 × 10^-5 K^-1 and α_Al = 2.2 × 10^-5K^-1)

1. 113.9 cm
2. 88 cm
3. 68 cm
4. 6.8 cm

Answer

68 cm

Reason — Given,

• Length of the copper rod, l_Cu = 88 cm
• α_Cu = 1.7 × 10^-5 K^-1, α_Al = 2.2 × 10^-5 K^-1

The difference in the lengths of the two rods will be independent of the temperature only if both the rods increase in length by exactly the same amount for any rise in temperature ΔT,

$\text l_{Cu}\space\alpha_{Cu}\space\Delta \text T = \text l_{Al}\space\alpha_{Al}\space\Delta \text T$

$\text l_{Al} = \dfrac{\text l_{Cu}\space\alpha_{Cu}}{\alpha_{Al}}$

Substituting the values,

$\text l_{Al} = \dfrac{88 \times (1.7 \times 10^{-5})}{2.2 \times 10^{-5}} \\[1em] = \dfrac{88 \times 1.7}{2.2} = \dfrac{149.6}{2.2} = 68\ \text{cm}$

A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be :

1. 450
2. 1000
3. 1800
4. 225

Answer

1800

Reason — Given,

• Radius of the black body, r₁ = 12 cm, temperature T₁ = 500 K, power P₁ = 450 W
• New radius, $\text r_2 = \dfrac{\text r_1}{2}$; new temperature, T₂ = 2T₁

By Stefan's law, the power radiated by a spherical black body of radius r is

$\text P = \sigma(4\pi \text r^2)\text T^4 \quad \Rightarrow \quad \text P \propto \text r^2\text T^4$

Therefore,

$\dfrac{\text P_2}{\text P_1} = \left(\dfrac{\text r_2}{\text r_1}\right)^2 \left(\dfrac{\text T_2}{\text T_1}\right)^4$

Substituting the values,

$\dfrac{\text P_2}{450} = \left(\dfrac{1}{2}\right)^2 \times (2)^4 = \dfrac{1}{4} \times 16 = 4$

$\text P_2 = 450 \times 4 = 1800\ \text W$

Two materials having coefficients of thermal conductivity 3K and K and thickness d and 3d respectively, are joined to form a slab as shown in figure. The temperature of the outer surfaces are θ₂ and θ₁ (θ₂ > θ₁). The temperature at the interface is :

1. $\dfrac{\theta_1 + \theta_2}{2}$

2. $\dfrac{\theta_1}{3} + \dfrac{2\theta_2}{3}$

3. $\dfrac{\theta_1}{6} + \dfrac{5\theta_2}{6}$

4. $\dfrac{\theta_1}{10} + \dfrac{9\theta_2}{10}$

Answer

$\dfrac{\theta_1}{10} + \dfrac{9\theta_2}{10}$

Reason — Given,

• First material : thermal conductivity 3K, thickness d, outer surface at θ₂
• Second material : thermal conductivity K, thickness 3d, outer surface at θ₁
• θ₂ > θ₁

Let θ be the temperature of the interface and A the area of cross-section.

The slabs are joined in series, so in the steady state the rate of flow of heat through both is the same,

$\dfrac{3\text{KA}(\theta_2 - \theta)}{\text d} = \dfrac{\text{KA}(\theta - \theta_1)}{3\text d}$

Cancelling K, A and d,

$9(\theta_2 - \theta) = \theta - \theta_1$

$9\theta_2 - 9\theta = \theta - \theta_1$

$10\theta = 9\theta_2 + \theta_1$

$\theta = \dfrac{9\theta_2 + \theta_1}{10} = \dfrac{\theta_1}{10} + \dfrac{9\theta_2}{10}$

A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K₁ and that of outer cylinder is K₂. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is :

1. $\dfrac{\text K_1 + \text K_2}{2}$

2. $\dfrac{\text K_1 + 3\text K_2}{4}$

3. $\dfrac{2\text K_1 + 3\text K_2}{5}$

4. $\text K_1 + \text K_2$

Answer

$\dfrac{\text K_1 + 3\text K_2}{4}$

Reason — Given,

• Inner cylinder : radius R, thermal conductivity K₁
• Outer shell : inner radius R and outer radius 2R, thermal conductivity K₂
• Heat flows along the length of the cylinder

Since the heat flows along the length, the inner cylinder and the outer shell have the same length and the same temperature difference across their ends. Hence they are joined in parallel.

The area of cross-section of the inner cylinder is

$\text A_1 = \pi \text R^2$

The area of cross-section of the outer shell is

$\text A_2 = \pi(2\text R)^2 - \pi \text R^2 = 4\pi \text R^2 - \pi \text R^2 = 3\pi \text R^2$

For conductors joined in parallel, the equivalent thermal conductivity is

$\text K = \dfrac{\text K_1\text A_1 + \text K_2\text A_2}{\text A_1 + \text A_2}$

Substituting the values,

$\text K = \dfrac{\text K_1(\pi \text R^2) + \text K_2(3\pi \text R^2)}{\pi \text R^2 + 3\pi \text R^2}$

$= \dfrac{\pi \text R^2(\text K_1 + 3\text K_2)}{4\pi \text R^2} = \dfrac{\text K_1 + 3\text K_2}{4}$

Temperature difference of 120°C is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ of same cross-section as AB and length $\dfrac{3\text L}{2}$ is connected across AB as shown in figure. In steady state, temperature difference between P and Q will be close to :

1. 45°C
2. 35°C
3. 75°C
4. 60°C

Answer

45°C

Reason — Given,

• Length of the rod AB = 2L, temperature difference between its ends A and B = 120°C
• The bent rod is PNMQ, of the same area of cross-section as AB and of total length $\dfrac{3\text L}{2}$
• From the figure, AP = QB = $\dfrac{\text L}{2}$, PQ = L, and in the bent rod PN = MQ = $\dfrac{\text L}{4}$, NM = L

Check on the length of the bent rod : Adding its three parts,

$\text{PN} + \text{NM} + \text{MQ} = \dfrac{\text L}{4} + \text L + \dfrac{\text L}{4} = \dfrac{3\text L}{2}$

which agrees with the length given in the question.

Thermal resistance : For a conductor of length l and area of cross-section A made of material of thermal conductivity K, the thermal resistance is

$\text R = \dfrac{\text l}{\text{KA}}$

Since K and A are the same throughout, the thermal resistance of any part is directly proportional to its length. Let R be the thermal resistance of a part of length L, that is,

$\text R = \dfrac{\text L}{\text{KA}}$

Then the resistance of a part of length $\dfrac{\text L}{2}$ is $\dfrac{\text R}{2}$, and that of a part of length $\dfrac{\text L}{4}$ is $\dfrac{\text R}{4}$.

Resistance of the bent rod PNMQ : The parts PN, NM and MQ carry the same heat one after the other, so they are in series and their resistances add,

$\text R_\text T = \dfrac{\text R}{4} + \text R + \dfrac{\text R}{4} = \dfrac{3\text R}{2}$

Resistance of the section PQ : Between the points P and Q, heat can travel by two separate paths — along the straight portion PQ of the rod AB, of resistance R, and along the bent rod PNMQ, of resistance $\dfrac{3\text R}{2}$. Both paths have the same temperature difference across them, so they are in parallel,

$\text R'_\text T = \dfrac{\text R \times \dfrac{3\text R}{2}}{\text R + \dfrac{3\text R}{2}}$

Simplifying the numerator and the denominator separately,

$\text R'_\text T = \dfrac{\dfrac{3\text R^2}{2}}{\dfrac{5\text R}{2}} = \dfrac{3\text R^2}{2} \times \dfrac{2}{5\text R}$

$\text R'_\text T = \dfrac{3\text R}{5}$

Total resistance of the network : The three sections AP, PQ and QB are joined end to end, so they are in series,

$\text R_{net} = \text R_{AP} + \text R'_\text T + \text R_{QB}$

$= \dfrac{\text R}{2} + \dfrac{3\text R}{5} + \dfrac{\text R}{2}$

$= \text R + \dfrac{3\text R}{5} = \dfrac{5\text R + 3\text R}{5} = \dfrac{8\text R}{5}$

Thermal current : In the steady state the same rate of flow of heat passes through the whole network,

$\text i = \dfrac{\text{Temperature difference between the ends A and B}}{\text R_{net}}$

$\text i = \dfrac{120}{\dfrac{8\text R}{5}} = \dfrac{120 \times 5}{8\text R} = \dfrac{75}{\text R}$

Temperature difference between P and Q : This thermal current flows through the section PQ, whose resistance is $\text R'_\text T$. Hence

$\Delta \theta_{PQ} = \text i \times \text R'_\text T = \dfrac{75}{\text R} \times \dfrac{3\text R}{5}$

$\Delta \theta_{PQ} = 45^\circ \text C$

The same result follows directly from the fact that in a series combination the temperature difference is shared in the ratio of the resistances,

$\Delta \theta_{PQ} = 120 \times \dfrac{\text R'_\text T}{\text R_{net}} = 120 \times \dfrac{3\text R/5}{8\text R/5} = 120 \times \dfrac{3}{8} = 45^\circ \text C$

The unit of thermal conductivity is :

1. J m^-1 K^-1
2. W m K^-1
3. W m^-1 K^-1
4. J m K^-1

Answer

W m^-1 K^-1

Reason — From the equation of heat conduction,

$\text K = \dfrac{\text Q\text l}{\text A(\theta_1 - \theta_2)\text t}$

the unit of K is

$\dfrac{\text{J} \times \text m}{\text m^2 \times \text K \times \text s} = \text{J s}^{-1}\text m^{-1}\text K^{-1}$

Since 1 J s^-1 = 1 W, the SI unit of thermal conductivity is W m^-1 K^-1.

An ice cube of dimensions 60 cm × 50 cm × 20 cm is placed in an insulation box of wall thickness 1 cm. The box keeping the ice cube at 0°C of temperature is brought to a room of temperature 40°C. The rate of melting of ice is approximately :

(Latent heat of fusion of ice is 3.4 × 10⁵ J kg^-1 and thermal conducting of insulation wall is 0.05 Wm^-1°C^-1)

1. 61 × 10^-3 kg s^-1
2. 61 × 10^-5 kg s^-1
3. 208 kg s^-1
4. 30 × 10^-5 kg s^-1

Answer

61 × 10^-5 kg s^-1

Reason — Given,

• Dimensions of the ice cube : 60 cm × 50 cm × 20 cm
• Thickness of the insulation wall, l = 1 cm = 0.01 m
• Temperature difference, θ₁ − θ₂ = 40 − 0 = 40°C
• Thermal conductivity, K = 0.05 W m^-1 °C^-1
• Latent heat of fusion of ice, L = 3.4 × 10⁵ J kg^-1

The total surface area of the box is

$\text A = 2\left[(60 \times 50) + (50 \times 20) + (60 \times 20)\right]\ \text{cm}^2$

$= 2(3000 + 1000 + 1200) = 10400\ \text{cm}^2 = 1.04\ \text m^2$

The rate of flow of heat into the box is

$\text H = \dfrac{\text{KA}(\theta_1 - \theta_2)}{\text l} = \dfrac{0.05 \times 1.04 \times 40}{0.01}$

$= \dfrac{2.08}{0.01} = 208\ \text{J s}^{-1}$

The rate of melting of ice is

$\dfrac{\text m}{\text t} = \dfrac{\text H}{\text L} = \dfrac{208}{3.4 \times 10^5} \\[1em] = 6.1 \times 10^{-4} = 61 \times 10^{-5}\ \text{kg s}^{-1}$

0.08 kg air is heated at constant volume through 5°C. The specific heat of air at constant volume is 0.17 kcal/kg°C and J = 4.18 joule/cal. The change in its internal energy is approximately :

1. 318 J
2. 298 J
3. 284 J
4. 142 J

Answer

284 J

Reason — Given,

• Mass of air, m = 0.08 kg
• Rise in temperature, ΔT = 5°C
• Specific heat at constant volume, c_v = 0.17 kcal kg^-1 °C^-1
• J = 4.18 J cal^-1

At constant volume no work is done by the gas, so by the first law of thermodynamics the whole of the heat supplied goes into increasing the internal energy,

$\Delta \text U = \text Q = \text m\text c_\text v\Delta \text T$

Substituting the values,

$\Delta \text U = 0.08 \times 0.17 \times 5 = 0.068\ \text{kcal} \\[1em] = 68\ \text{cal}$

Converting into joule,

$\Delta \text U = 68 \times 4.18 = 284.2\ \text J$

The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are 8 Ω and 10 Ω respectively. After inserting in a hot bath of temperature 400°C, the resistance of platinum wire is :

1. 2 Ω
2. 16 Ω
3. 8 Ω
4. 10 Ω

Answer

16 Ω

Reason — Given,

• Resistance at the ice point, R₀ = 8 Ω
• Resistance at the steam point, R₁₀₀ = 10 Ω
• Temperature of the hot bath, t = 400°C

For a platinum resistance thermometer, the temperature is given by

$\text t = \dfrac{\text R_\text t - \text R_0}{\text R_{100} - \text R_0} \times 100$

Substituting the values,

$400 = \dfrac{\text R_\text t - 8}{10 - 8} \times 100$

$400 = \dfrac{(\text R_\text t - 8) \times 100}{2} = 50(\text R_\text t - 8)$

$\text R_\text t - 8 = 8 \quad \Rightarrow \quad \text R_\text t = 16\ \Omega$

On Celsius scale the temperature of body increases by 40°C. The increase in temperature on Fahrenheit scale is :

1. 70°F
2. 68°F
3. 72°F
4. 75°F

Answer

72°F

Reason — Given,

• Increase in temperature on the Celsius scale, ΔC = 40°C

The interval between the ice point and the steam point contains 100 divisions on the Celsius scale and 180 divisions on the Fahrenheit scale. Hence, for a temperature difference,

$\dfrac{\Delta \text C}{100} = \dfrac{\Delta \text F}{180} \quad \Rightarrow \quad \Delta \text F = \dfrac{9}{5}\Delta \text C$

Substituting the value,

$\Delta \text F = \dfrac{9}{5} \times 40 = 72^\circ \text F$

The factor 32 is not added here, because it applies only to the conversion of a temperature reading and not of a temperature difference.

A metallic bar of Young's modulus, 0.5 × 10¹¹ Nm^-2 and coefficient of linear expansion 10^-5 °C^-1, length 1 m and area of cross-section 10^-3 m² is heated from 0°C to 100°C without expansion or bending. The compressive force developed in it :

1. 5 × 10³ N
2. 50 × 10³ N
3. 100 × 10³ N
4. 2 × 10³ N

Answer

50 × 10³ N

Reason — Given,

• Young's modulus, Y = 0.5 × 10¹¹ N m^-2
• Coefficient of linear expansion, α = 10^-5 °C^-1
• Area of cross-section, A = 10^-3 m²
• Rise in temperature, ΔT = 100 − 0 = 100°C

On heating, the bar tends to expand by Δl = l α ΔT, but the expansion is prevented. The bar therefore behaves as though it had been compressed by this amount, and a compressive force is developed in it. From the definition of Young's modulus,

$\text Y = \dfrac{\text F/\text A}{\Delta \text l/\text l} \quad \Rightarrow \quad \text F = \text{YA}\dfrac{\Delta \text l}{\text l}$

Putting Δl = l α ΔT,

$\text F = \text{YA}\alpha\Delta \text T$

Substituting the values,

$\text F = (0.5 \times 10^{11}) \times 10^{-3} \times 10^{-5} \times 100 \\[1em] = (5 \times 10^7) \times 10^{-5} \times 100 \\[1em] = 5 \times 10^4 = 50 \times 10^3\ \text N$

Note that the force does not depend on the length of the bar.

Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity 2K while that in the middle has thermal conductivity K. The left end of the combination is maintained at temperature 3T and the right end at T. The rods are thermally insulated from outside. In steady state, temperature at the left junction is T₁ and that at the right junction is T₂. The ratio T₁/T₂ is :

1. 3/2
2. 4/3
3. 5/3
4. 5/4

Answer

5/3

Reason — Given,

• Three identical rods of the same length l and area of cross-section A, joined in series
• Thermal conductivities : 2K, K and 2K respectively
• Left end at 3T and right end at T
• Temperature of the left junction = T₁, of the right junction = T₂

In the steady state the rate of flow of heat through all the three rods is the same,

$\dfrac{2\text{KA}(3\text T - \text T_1)}{\text l} = \dfrac{\text{KA}(\text T_1 - \text T_2)}{\text l} = \dfrac{2\text{KA}(\text T_2 - \text T)}{\text l}$

Cancelling K, A and l,

$2(3\text T - \text T_1) = (\text T_1 - \text T_2) = 2(\text T_2 - \text T)$

From the first and the third terms :

$2(3\text T - \text T_1) = 2(\text T_2 - \text T)$

$3\text T - \text T_1 = \text T_2 - \text T \quad \Rightarrow \quad \text T_1 + \text T_2 = 4\text T \qquad \ldots(\text i)$

From the first and the second terms :

$2(3\text T - \text T_1) = \text T_1 - \text T_2$

$6\text T - 2\text T_1 = \text T_1 - \text T_2 \quad \Rightarrow \quad 3\text T_1 - \text T_2 = 6\text T \qquad \ldots(\text{ii})$

From equation (i), T₂ = 4T − T₁. Substituting in equation (ii),

$3\text T_1 - (4\text T - \text T_1) = 6\text T$

$4\text T_1 = 10\text T \quad \Rightarrow \quad \text T_1 = \dfrac{5\text T}{2}$

$\text T_2 = 4\text T - \dfrac{5\text T}{2} = \dfrac{3\text T}{2}$

Therefore,

$\dfrac{\text T_1}{\text T_2} = \dfrac{5\text T/2}{3\text T/2} = \dfrac{5}{3}$

A bimetallic strip is formed out of two identical strips, one of copper and the other of brass. The coefficients of linear expansion of the two metals are α_C and α_B. On heating, the temperature of the strip goes up by ΔT and the strip bends to form an arc of radius of curvature R. Then R is :

1. proportional to ΔT
2. inversely proportional to ΔT
3. proportional to | α_B − α_C|
4. inversely proportional to | α_B − α_C|

Answer

2. inversely proportional to ΔT

4. inversely proportional to | α_B − α_C|

Reason — Given,

• Two identical strips, one of copper and one of brass, of coefficients of linear expansion α_C and α_B
• Rise in temperature = ΔT
• Radius of curvature of the arc formed = R

Let l₀ be the original length of each strip and t the thickness of the bimetallic strip.

Expansion of the two strips : On heating, each strip tends to expand by an amount decided by its own coefficient of linear expansion, so their lengths become

$\text l_\text B = \text l_0(1 + \alpha_\text B\Delta \text T) \quad \text{and} \quad \text l_\text C = \text l_0(1 + \alpha_\text C\Delta \text T)$

Since brass expands more than copper (α_B > α_C), the brass strip becomes longer than the copper strip. The two strips are firmly rivetted together and cannot slide over each other, so the strip bends with brass on the outside of the curve and copper on the inside.

Geometry of the bent strip : Let R be the radius of curvature of the interface between the two strips and θ the angle subtended by the arc at the centre. The brass strip then lies along an arc of radius $\left(\text R + \dfrac{\text t}{2}\right)$ and the copper strip along an arc of radius $\left(\text R - \dfrac{\text t}{2}\right)$.

Using the relation arc = radius × angle for each strip,

$\left(\text R + \dfrac{\text t}{2}\right)\theta = \text l_0(1 + \alpha_\text B\Delta \text T) \qquad \ldots(\text i)$

$\left(\text R - \dfrac{\text t}{2}\right)\theta = \text l_0(1 + \alpha_\text C\Delta \text T) \qquad \ldots(\text{ii})$

Eliminating l₀ and θ : Dividing equation (i) by equation (ii), both l₀ and θ cancel out,

$\dfrac{\text R + \dfrac{\text t}{2}}{\text R - \dfrac{\text t}{2}} = \dfrac{1 + \alpha_\text B\Delta \text T}{1 + \alpha_\text C\Delta \text T}$

Cross-multiplying,

$\left(\text R + \dfrac{\text t}{2}\right)(1 + \alpha_\text C\Delta \text T) = \left(\text R - \dfrac{\text t}{2}\right)(1 + \alpha_\text B\Delta \text T)$

Expanding both sides,

$\text R + \text R\alpha_\text C\Delta \text T + \dfrac{\text t}{2} + \dfrac{\text t}{2}\alpha_\text C\Delta \text T = \text R + \text R\alpha_\text B\Delta \text T - \dfrac{\text t}{2} - \dfrac{\text t}{2}\alpha_\text B\Delta \text T$

The term R cancels from both sides. The terms $\dfrac{\text t}{2}\alpha_\text C\Delta \text T$ and $\dfrac{\text t}{2}\alpha_\text B\Delta \text T$ are products of two very small quantities and are neglected. Then

$\text R\alpha_\text C\Delta \text T + \dfrac{\text t}{2} = \text R\alpha_\text B\Delta \text T - \dfrac{\text t}{2}$

Collecting the t terms on one side and the R terms on the other,

$\dfrac{\text t}{2} + \dfrac{\text t}{2} = \text R\alpha_\text B\Delta \text T - \text R\alpha_\text C\Delta \text T$

$\text t = \text R(\alpha_\text B - \alpha_\text C)\Delta \text T$

$\text R = \dfrac{\text t}{(\alpha_\text B - \alpha_\text C)\Delta \text T}$

Conclusion : In this expression the thickness t is a constant for a given strip. Hence

$\text R \propto \dfrac{1}{\Delta \text T} \quad \text{and} \quad \text R \propto \dfrac{1}{|\alpha_\text B - \alpha_\text C|}$

So the radius of curvature is inversely proportional to the rise in temperature and inversely proportional to the difference of the coefficients of linear expansion. Options 2 and 4 are correct, and options 1 and 3 are incorrect.

Physically, a greater rise in temperature or a greater difference in the coefficients of expansion makes the strip bend more sharply, and a sharper bend means a smaller radius of curvature. This is why a bimetallic strip is used as the sensitive element in an electric thermostat.

Select the correct alternative (s) :

Two bodies A and B, having same outer surface areas, have thermal emissivities of 0.01 and 0.81 respectively. They emit total radiant energy at the same rate. The wavelength λ_B corresponding to maximum spectral radiancy in the radiation from B is displaced from the wavelength corresponding to maximum spectral radiancy in the radiation from A, by 1.00 μm. If the temperature of A is 5802 K, then :

1. the temperature of B is 1934 K
2. λ_B = 1.5 μm
3. the temperature of B is 11604 K
4. the temperature of B is 2901 K

Answer

1. the temperature of B is 1934 K

2. λ_B = 1.5 μm

Reason — Given,

• Emissivities, e_A = 0.01 and e_B = 0.81
• Both bodies have the same outer surface area A and emit total radiant energy at the same rate
• Temperature of A, T_A = 5802 K
• λ_B − λ_A = 1.00 μm

Temperature of B : Since the rates of emission are equal,

$\text e_\text A\sigma\text A\text T_\text A^4 = \text e_\text B\sigma\text A\text T_\text B^4$

$0.01\text T_\text A^4 = 0.81\text T_\text B^4$

$\left(\dfrac{\text T_\text A}{\text T_\text B}\right)^4 = \dfrac{0.81}{0.01} = 81 \quad \Rightarrow \quad \dfrac{\text T_\text A}{\text T_\text B} = 3$

$\text T_\text B = \dfrac{5802}{3} = 1934\ \text K$

Hence option 1 is correct, and options 3 and 4 are incorrect.

Wavelength λ_B : By Wien's displacement law, λ_mT = b, where b = 2.9 × 10^-3 m K.

$\lambda_\text A = \dfrac{2.9 \times 10^{-3}}{5802} = 5.0 \times 10^{-7}\ \text m = 0.5\ \mu\text m$

$\lambda_\text B = \dfrac{2.9 \times 10^{-3}}{1934} = 1.5 \times 10^{-6}\ \text m = 1.5\ \mu\text m$

Hence option 2 is correct. The displacement is λ_B − λ_A = 1.5 − 0.5 = 1.00 μm, which agrees with the data given in the question.

A human body has a surface area of approximately 1 m². The normal body temperature is 10 K above the surrounding room temperature T₀. Take the room temperature to be T₀ = 300 K. For T₀ = 300 K, the value of σ T₀⁴ = 460 Wm^-2 (where σ is the Stefan-Boltzmann constant). Which of the following option(s) is are correct?

1. If the surrounding temperature reduces by a small amount ΔT₀ << T₀, then to maintain the same body temperature the same (living) human being needs to radiate ΔW = 4 σT₀³ΔT₀ more energy per unit time
2. Reducing the exposed surface area of the body (e.g. by curling up) allows humans to maintain the same body temperature while reducing the energy lost by radiation
3. If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths
4. The amount of energy radiated by the body in 1 second is close to 60 joules.

Answer

1. If the surrounding temperature reduces by a small amount ΔT₀ << T₀, then to maintain the same body temperature the same (living) human being needs to radiate ΔW = 4 σT₀³ΔT₀ more energy per unit time

2. Reducing the exposed surface area of the body (e.g. by curling up) allows humans to maintain the same body temperature while reducing the energy lost by radiation

4. The amount of energy radiated by the body in 1 second is close to 60 joules

Reason — Given,

• Surface area of the body, A = 1 m²
• Body temperature, T = T₀ + 10, with T₀ = 300 K
• σT₀⁴ = 460 W m^-2

The net energy radiated per second by the body is

$\text W = \sigma \text A(\text T^4 - \text T_0^4)$

Option 1 : If the surrounding temperature falls by ΔT₀ while the body temperature is kept the same, the extra energy that must be radiated per unit time is obtained by differentiating σT₀⁴ with respect to T₀,

$\Delta \text W = \left|\dfrac{\text d}{\text{dT}_0}(\sigma \text T_0^4)\right|\Delta \text T_0 = 4\sigma \text T_0^3\Delta \text T_0$

Hence option 1 is correct.

Option 2 : Since W = σA(T⁴ − T₀⁴) is directly proportional to the exposed surface area A, reducing A by curling up reduces the energy lost by radiation while the body temperature stays the same. Hence option 2 is correct.

Option 3 : By Wien's displacement law, $\lambda_\text m = \dfrac{\text b}{\text T}$, so λ_m is inversely proportional to T. If the body temperature rises, the peak shifts towards shorter wavelengths, not longer. Hence option 3 is incorrect.

Option 4 : Substituting T = T₀ + 10,

$\text W = \sigma \text A\text T_0^4\left[\left(1 + \dfrac{10}{\text T_0}\right)^4 - 1\right]$

Since $\dfrac{10}{\text T_0}$ is small, expanding by the binomial theorem and keeping only the first-order term,

$\text W \approx \sigma \text A\text T_0^4 \times \dfrac{4 \times 10}{\text T_0} = 460 \times 1 \times \dfrac{40}{300}$

$= 61.3\ \text J\text s^{-1} \approx 60\ \text J\text s^{-1}$

Hence option 4 is correct.

The specific heat capacity of a substance is temperature dependent and is given by the formula C = kT, where k is a constant of suitable dimensions in SI units, and T is the absolute temperature. If the heat required to raise the temperature of 1 kg of the substance from −73°C to 27°C is nk, the value of n is ............... .

[Given: 0 K = −273 °C.]

Answer

Given,

• Specific heat capacity, C = kT
• Mass of the substance, m = 1 kg
• Initial temperature, t₁ = − 73°C = − 73 + 273 = 200 K
• Final temperature, t₂ = 27°C = 27 + 273 = 300 K
• Heat required = nk

Since the specific heat capacity varies with temperature, the heat required cannot be found by simple multiplication; it must be obtained by integration.

For an infinitesimally small rise in temperature dT, the heat required is

$\text{dQ} = \text m\text C\text{dT} = \text m(\text{kT})\text{dT}$

Integrating between the limits T = 200 K and T = 300 K,

$\text Q = \int_{200}^{300} \text m\text k\text T\text{dT}$

Putting m = 1 kg and taking the constant k outside,

$\text Q = \text k\int_{200}^{300} \text T\text{dT} = \text k\left[\dfrac{\text T^2}{2}\right]_{200}^{300}$

Evaluating the limits,

$\text Q = \dfrac{\text k}{2}\left[(300)^2 - (200)^2\right]$

$= \dfrac{\text k}{2}\left[90000 - 40000\right] = \dfrac{\text k}{2} \times 50000$

$\text Q = 25000\text k$

Comparing with Q = nk,

$\text n = 25000$

Hence, the value of n is 25000.

Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T₁ = 300 K and T₂ = 100 K, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K₁ and K₂ respectively. If the temperature at the junction of the two cylinders in the steady state is 200 K, then K₁/K₂ = ............... . Round off up to two decimal places.

Answer

Given,

• Both cylinders have the same length l
• Radius of the smaller cylinder = r, radius of the bigger cylinder = 2r
• Thermal conductivity of the smaller cylinder = K₁, of the bigger cylinder = K₂
• T₁ = 300 K, T₂ = 100 K, junction temperature T = 200 K

The areas of cross-section of the two cylinders are

$\text A_1 = \pi \text r^2 \quad \text{and} \quad \text A_2 = \pi(2\text r)^2 = 4\pi \text r^2$

The cylinders are connected in series, so in the steady state the rate of flow of heat through both is the same,

$\dfrac{\text K_1\text A_1(\text T_1 - \text T)}{\text l} = \dfrac{\text K_2\text A_2(\text T - \text T_2)}{\text l}$

Substituting the values,

$\text K_1(\pi \text r^2)(300 - 200) = \text K_2(4\pi \text r^2)(200 - 100)$

Cancelling πr²,

$\text K_1 \times 100 = \text K_2 \times 4 \times 100$

$\text K_1 = 4\text K_2$

$\dfrac{\text K_1}{\text K_2} = 4.00$

Hence, the value of K₁/K₂ is 4.00.

The left and right compartments of a thermally isolated container of length L are separated by a thermally conducting, movable piston of area A. The left and right compartments are filled with $\dfrac{3}{2}$ and 1 mole of an ideal gas, respectively. In the left compartment the piston is attached by a spring with spring constant k and natural length $\dfrac{2\text L}{5}$. In thermodynamic equilibrium, the piston is at a distance $\dfrac{\text L}{2}$ from the left and right edges of the container as shown in the figure. Under the above conditions, if the pressure in the right compartment is $\text P = \dfrac{\text{kL}}{\text A}\alpha$, then the value of α is ............... .

Answer

Given,

• Length of the container = L, area of the piston = A
• Left compartment : $\dfrac{3}{2}$ mole of an ideal gas; right compartment : 1 mole of an ideal gas
• Spring constant = k, natural length of the spring = $\dfrac{2\text L}{5}$
• In equilibrium, the piston is at a distance $\dfrac{\text L}{2}$ from each edge

Temperatures and volumes : The piston is thermally conducting and the container is thermally isolated, so in thermodynamic equilibrium the gases in both the compartments are at the same temperature T. The volume of each compartment is

$\text V = \text A \times \dfrac{\text L}{2} = \dfrac{\text{AL}}{2}$

Pressures of the two gases : From the ideal gas equation PV = nRT, at the same temperature and the same volume the pressure is proportional to the number of moles. Hence

$\text P_{left} = \dfrac{(3/2)\text{RT}}{\text{AL}/2} \quad \text{and} \quad \text P_{right} = \dfrac{\text{RT}}{\text{AL}/2}$

$\dfrac{\text P_{left}}{\text P_{right}} = \dfrac{3}{2} \quad \Rightarrow \quad \text P_{left} = \dfrac{3}{2}\text P_{right}$

Spring force : The spring lies in the left compartment. Its present length is $\dfrac{\text L}{2}$ while its natural length is $\dfrac{2\text L}{5}$, so it is stretched by

$\text x = \dfrac{\text L}{2} - \dfrac{2\text L}{5} = \dfrac{5\text L - 4\text L}{10} = \dfrac{\text L}{10}$

Since the spring is stretched, it pulls the piston towards the left wall with a force

$\text F = \text{kx} = \dfrac{\text{kL}}{10}$

Force balance on the piston : The gas in the left compartment pushes the piston to the right with force P_leftA, while the gas in the right compartment pushes it to the left with force P_rightA, and the stretched spring also pulls it to the left with force F. In equilibrium,

$\text P_{left}\text A = \text P_{right}\text A + \dfrac{\text{kL}}{10}$

Substituting $\text P_{left} = \dfrac{3}{2}\text P_{right}$,

$\dfrac{3}{2}\text P_{right}\text A - \text P_{right}\text A = \dfrac{\text{kL}}{10}$

$\dfrac{1}{2}\text P_{right}\text A = \dfrac{\text{kL}}{10}$

$\text P_{right} = \dfrac{\text{kL}}{5\text A} = \dfrac{\text{kL}}{\text A} \times \dfrac{1}{5}$

Comparing with $\text P = \dfrac{\text{kL}}{\text A}\alpha$,

$\alpha = \dfrac{1}{5} = 0.2$

Hence, the value of α is 0.2.

Two identical plates P and Q, radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures T_P and T_Q, respectively, with T_Q < T_P, as shown in Fig. 1. The radiated power transferred per unit area from P to Q is W₀. Subsequently, two more plates, identical to P and Q, are introduced between P and Q, as shown in Fig. 2. Assume that heat transfer takes place only between adjacent plates. If the power transferred per unit area in the direction from P to Q (Fig. 2) in the steady state is W_S, then the ratio $\dfrac{\text W_0}{\text W_\text S}$ is ............... .

Answer

Given,

• Plates P and Q radiate as perfect black bodies, at temperatures T_P and T_Q, with T_Q < T_P
• In Fig. 1 the power transferred per unit area from P to Q is W₀
• In Fig. 2, two more identical plates are introduced between P and Q
• Heat transfer takes place only between adjacent plates

Case 1 (Fig. 1) : There is a single gap between P and Q. By Stefan's law, the net power transferred per unit area is

$\text W_0 = \sigma(\text T_\text P^4 - \text T_\text Q^4) \qquad \ldots(\text i)$

Case 2 (Fig. 2) : Now there are four plates in all — P, the two intermediate plates at steady temperatures T₁ and T₂, and Q. Hence there are three gaps, and in the steady state the same power W_S must be transferred across each gap, since otherwise the intermediate plates would go on heating up or cooling down.

Therefore,

$\text W_\text S = \sigma(\text T_\text P^4 - \text T_1^4) = \sigma(\text T_1^4 - \text T_2^4) = \sigma(\text T_2^4 - \text T_\text Q^4)$

Adding the three expressions,

$3\text W_\text S = \sigma\left[(\text T_\text P^4 - \text T_1^4) + (\text T_1^4 - \text T_2^4) + (\text T_2^4 - \text T_\text Q^4)\right]$

The intermediate terms cancel out, giving

$3\text W_\text S = \sigma(\text T_\text P^4 - \text T_\text Q^4) \qquad \ldots(\text{ii})$

Comparing equations (i) and (ii),

$3\text W_\text S = \text W_0$

$\dfrac{\text W_0}{\text W_\text S} = 3$

Hence, the required ratio is 3.