Thermal Properties of Matter — NCERT Exercises

The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on Celsius and Fahrenheit scales.

Answer

Given,

• Triple point of neon, T = 24.57 K
• Triple point of carbon dioxide, T = 216.55 K

The Kelvin and the Celsius scales are related as

$\text t_\text C = \text T - 273.15$

and the Celsius and the Fahrenheit scales are related as

$\text t_\text F = \dfrac{9}{5}\text t_\text C + 32$

For neon :

$\text t_\text C = 24.57 - 273.15 = -248.58^\circ \text C$

$\text t_\text F = \dfrac{9}{5} \times (-248.58) + 32 \\[1em] = -447.44 + 32 = -415.44^\circ \text F$

For carbon dioxide :

$\text t_\text C = 216.55 - 273.15 = -56.60^\circ \text C$

$\text t_\text F = \dfrac{9}{5} \times (-56.60) + 32 \\[1em] = -101.88 + 32 = -69.88^\circ \text F$

Hence, the triple point of neon is − 248.58°C or − 415.44°F, and that of carbon dioxide is − 56.60°C or − 69.88°F.

The triple point of water defined to be T_A = 200 A on an absolute scale A and T_B = 350 B on an absolute scale B. Find the relation between T_A and T_B.

Answer

Given,

• Triple point of water on scale A, T_A = 200 A
• Triple point of water on scale B, T_B = 350 B

Both A and B are absolute scales, so on each of them the zero of the scale is absolute zero. The triple point of water is a single physical state, and therefore the readings 200 A and 350 B represent the same temperature.

If the triple point of water is assigned the value T_tr on the Kelvin scale, then for any absolute scale the temperature of a body is proportional to its reading. Hence, for the triple point,

$\dfrac{\text T_\text A}{200} = \dfrac{\text T_\text B}{350}$

Therefore, for any temperature measured on the two scales,

$\text T_\text A = \dfrac{200}{350}\text T_\text B = \dfrac{4}{7}\text T_\text B$

Hence, the required relation is $\text T_\text A = \dfrac{4}{7}\text T_\text B$, that is, 7 T_A = 4 T_B.

The electrical resistance (in ohms) of a certain thermometer varies with temperature roughly as :

R_T = R₀[1 + 0.005 (T − T₀)].

The resistance is 101.6 Ω at the triple point of water (273.16 K) and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?

Answer

Given,

• Resistance at the triple point of water, R₁ = 101.6 Ω at T₁ = 273.16 K
• Resistance at the melting point of lead, R₂ = 165.5 Ω at T₂ = 600.5 K
• Resistance at the unknown temperature, R = 123.4 Ω

The resistance varies linearly with temperature as

$\text R_\text T = \text R_0[1 + \alpha(\text T - \text T_0)]$

Since the relation is linear, the change in resistance is directly proportional to the change in temperature. Therefore,

$\dfrac{\text R - \text R_1}{\text R_2 - \text R_1} = \dfrac{\text T - \text T_1}{\text T_2 - \text T_1}$

Substituting the given values,

$\dfrac{123.4 - 101.6}{165.5 - 101.6} = \dfrac{\text T - 273.16}{600.5 - 273.16}$

$\dfrac{21.8}{63.9} = \dfrac{\text T - 273.16}{327.34}$

$\text T - 273.16 = 0.3412 \times 327.34 = 111.68$

$\text T = 273.16 + 111.68 = 384.84\ \text K$

Hence, the temperature at which the resistance is 123.4 Ω is 384.84 K.

Answer the following :

(a) The triple point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?

(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?

(c) The absolute temperature (Kelvin scale) T is related to the temperature t_c on the Celsius scale by

t_c = T − 273.15

Why do we have 273.15 in this relation, and not 273.16?

(d) What is the temperature of the triple point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Answer

(a) The triple point of water is that unique combination of temperature and pressure at which the solid, the liquid and the vapour phases of water co-exist in stable equilibrium. Since this state occurs at only one value of temperature and one value of pressure, it is perfectly reproducible everywhere.

The melting point of ice and the boiling point of water, on the other hand, depend upon the external pressure. They change if the pressure changes, and so they are not unique. That is why they are unsuitable as standard fixed points, while the triple point of water is taken as the standard fixed point in modern thermometry.

(b) The other fixed point on the Kelvin scale is absolute zero, which is assigned the value 0 K. It is the lowest possible temperature, at which molecular motion would stop completely.

(c) On the Celsius scale the ice point (the melting point of ice at standard atmospheric pressure) is taken as 0°C, and this corresponds to 273.15 K on the Kelvin scale. The triple point of water is not the ice point; it lies 0.01°C above it, that is, at 273.16 K.

Since the relation t_c = T − 273.15 converts a Kelvin reading into a Celsius reading, it must use the Kelvin value of the ice point. Hence 273.15 appears in the relation and not 273.16.

(d) On the Kelvin absolute scale, the triple point of water is 273.16 K and the interval between the ice point and the steam point contains 100 divisions. On the Fahrenheit scale the same interval contains 180 divisions. Hence, one Kelvin division corresponds to $\dfrac{180}{100} = \dfrac{9}{5}$ Fahrenheit divisions.

For an absolute scale whose unit interval size equals that of the Fahrenheit scale,

$\text T_\text F = \dfrac{9}{5} \times 273.16 = 491.69$

Hence, the triple point of water on this absolute scale is 491.69 units.

Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?

(b) What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

Answer

Given,

• For thermometer A : P_tr = 1.250 × 10⁵ Pa, P = 1.797 × 10⁵ Pa
• For thermometer B : P_tr = 0.200 × 10⁵ Pa, P = 0.287 × 10⁵ Pa
• Triple point of water, T_tr = 273.16 K

(a) In a constant-volume gas thermometer the volume is kept fixed, so the pressure of the gas is directly proportional to its absolute temperature, that is, P ∝ T. Hence

$\text T = \text T_{tr} \times \dfrac{\text P}{\text P_{tr}}$

For thermometer A :

$\text T = 273.16 \times \dfrac{1.797 \times 10^5}{1.250 \times 10^5} \\[1em] = 273.16 \times 1.4376 = 392.69\ \text K$

For thermometer B :

$\text T = 273.16 \times \dfrac{0.287 \times 10^5}{0.200 \times 10^5} \\[1em] = 273.16 \times 1.435 = 391.98\ \text K$

Hence, the melting point of sulphur is 392.69 K as read by thermometer A and 391.98 K as read by thermometer B.

(b) The two readings differ slightly because oxygen and hydrogen are not perfectly ideal gases. The relation P ∝ T holds exactly only for an ideal gas, and a real gas departs from ideal behaviour, the departure being different for different gases.

The results depend somewhat on the choice of gas, but the less dense the gas in the bulb, the better the results for different gases agree. Hence, to reduce the discrepancy the experiment should be repeated with smaller and smaller quantities of the gases, that is, at lower and lower pressures, and the readings should be extrapolated to zero pressure. At zero pressure both the gases behave as ideal gases and the two thermometers give the same reading.

A steel tape 1 m long is correctly calibrated for a temperature of 27°C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45°C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27°C? The coefficient of linear expansion of steel is 1.2 × 10^-5 °C^-1.

Answer

Given,

• Reading of the tape at 45°C, l = 63.0 cm
• Temperature of calibration = 27°C
• Temperature on the hot day = 45°C, so Δt = 45 − 27 = 18°C
• Coefficient of linear expansion of steel, α = 1.2 × 10^-5 °C^-1

The tape is correctly calibrated at 27°C. On the hot day the tape itself expands, so each centimetre division of the tape becomes longer than a true centimetre. The true length of one division of the tape at 45°C is

$1 \times (1 + \alpha\Delta \text t) = 1 + (1.2 \times 10^{-5} \times 18) \\[1em] = 1 + 2.16 \times 10^{-4}\ \text{cm}$

Since the tape shows a reading of 63.0 cm, the actual length of the rod at 45°C is

$\text l_{45} = 63.0 \times (1 + 2.16 \times 10^{-4}) \\[1em] = 63.0 + 0.0136 = 63.0136\ \text{cm}$

The rod is also made of steel, so on cooling from 45°C to 27°C its length becomes

$\text l_{27} = \dfrac{\text l_{45}}{1 + \alpha\Delta \text t} = \dfrac{63.0136}{1 + 2.16 \times 10^{-4}} \\[1em] = 63.0\ \text{cm}$

Hence, the actual length of the steel rod on the hot day is 63.0136 cm, and at 27°C it is 63.0 cm.

At 27°C the tape is correctly calibrated, so the reading of the tape is itself the true length of the rod.

A large steel wheel is to be fitted on to a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using 'dry ice'. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: α_Steel = 1.20 × 10^-5 K^-1.

Answer

Given,

• Diameter of the shaft at 27°C, d = 8.70 cm
• Diameter of the hole in the wheel, d' = 8.69 cm
• Coefficient of linear expansion of steel, α = 1.20 × 10^-5 K^-1
• Initial temperature, T₁ = 27°C

The wheel will just slip on the shaft when, on cooling, the diameter of the shaft becomes equal to the diameter of the hole in the wheel. Hence the required change in diameter is

$\Delta \text d = 8.69 - 8.70 = -0.01\ \text{cm}$

For linear expansion,

$\Delta \text d = \text d\alpha\Delta \text T$

$\Delta \text T = \dfrac{\Delta \text d}{\text d\alpha} = \dfrac{-0.01}{8.70 \times (1.20 \times 10^{-5})} \\[1em] = \dfrac{-0.01}{1.044 \times 10^{-4}} = -95.8^\circ \text C$

The final temperature of the shaft is

$\text T_2 = \text T_1 + \Delta \text T = 27 - 95.8 = -68.8^\circ \text C$

Hence, the wheel will slip on the shaft when the shaft is cooled to about − 68.8°C.

A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C? The coefficient of linear expansion of copper is 1.70 × 10^-5°C^-1.

Answer

Given,

• Diameter of the hole at 27°C, d = 4.24 cm
• Rise in temperature, Δt = 227 − 27 = 200°C
• Coefficient of linear expansion of copper, α = 1.70 × 10^-5 °C^-1

On heating, the hole in the sheet expands in exactly the same way as a disc of copper filling the hole would expand. Hence the change in the diameter of the hole is

$\Delta \text d = \text d\alpha\Delta \text t$

Substituting the given values,

$\Delta \text d = 4.24 \times (1.70 \times 10^{-5}) \times 200 \\[1em] = 4.24 \times 3.4 \times 10^{-3} \\[1em] = 1.44 \times 10^{-2}\ \text{cm}$

Hence, the diameter of the hole increases by 1.44 × 10^-2 cm.

A brass wire 1.8 m long at 27°C is held taut with negligible tension between two rigid supports. If the wire is cooled to a temperature of − 39°C, what tension is developed in the wire, if its diameter is 2.0 mm? The coefficient of linear expansion of brass is 2.0 × 10^-5 °C^-1 and the Young's modulus is 0.91 × 10¹¹ N m^-2 (or Pa).

Answer

Given,

• Length of the brass wire, l = 1.8 m
• Fall in temperature, Δt = 27 − (− 39) = 66°C
• Diameter of the wire = 2.0 mm, so radius r = 1.0 mm = 1.0 × 10^-3 m
• Coefficient of linear expansion of brass, α = 2.0 × 10^-5 °C^-1
• Young's modulus, Y = 0.91 × 10¹¹ N m^-2

The area of cross-section of the wire is

$\text A = \pi \text r^2 = 3.14 \times (1.0 \times 10^{-3})^2 \\[1em] = 3.14 \times 10^{-6}\ \text m^2$

On cooling, the wire tends to contract by Δl = l α Δt, but the rigid supports do not allow it to contract. The wire therefore behaves as though it has been stretched by this amount, and a tension is developed in it. From the definition of Young's modulus,

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

Putting Δl = l α Δt,

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

Substituting the values,

$\text F = (0.91 \times 10^{11}) \times (3.14 \times 10^{-6}) \times (2.0 \times 10^{-5}) \times 66 \\[1em] = 2.857 \times 10^5 \times 2.0 \times 10^{-5} \times 66 \\[1em] = 5.714 \times 66 = 377.1\ \text N$

Hence, the tension developed in the wire is about 3.8 × 10² N.

A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0°C? Is there a 'thermal stress' developed at the junction? The ends of the rod are free to expand (Coefficient of linear expansion of brass = 2.0 × 10^-5 K^-1, steel = 1.2 × 10^-5 K^-1).

Answer

Given,

• Length of each rod, l = 50 cm = 0.50 m
• Rise in temperature, Δt = 250 − 40 = 210°C
• Coefficient of linear expansion of brass, α_b = 2.0 × 10^-5 K^-1
• Coefficient of linear expansion of steel, α_s = 1.2 × 10^-5 K^-1

Since the ends are free to expand, each rod expands independently.

The increase in the length of the brass rod is

$\Delta \text l_\text b = \text l\alpha_\text b\Delta \text t = 0.50 \times (2.0 \times 10^{-5}) \times 210 \\[1em] = 2.1 \times 10^{-3}\ \text m$

The increase in the length of the steel rod is

$\Delta \text l_\text s = \text l\alpha_\text s\Delta \text t = 0.50 \times (1.2 \times 10^{-5}) \times 210 \\[1em] = 1.26 \times 10^{-3}\ \text m$

The change in the length of the combined rod is

$\Delta \text l = \Delta \text l_\text b + \Delta \text l_\text s \\[1em] = (2.1 \times 10^{-3}) + (1.26 \times 10^{-3}) \\[1em] = 3.36 \times 10^{-3}\ \text m = 0.336\ \text{cm}$

Hence, the length of the combined rod increases by 0.336 cm.

No thermal stress is developed at the junction. Thermal stress arises only when the expansion of a rod is prevented. Here the ends of the combined rod are free to expand, so each rod expands freely by its own amount and no restoring force is set up at the junction.

The coefficient of volume expansion of glycerine is 49 × 10^-5 °C^-1. Find the fractional change in its density for a 30°C rise in temperature.

Answer

Given,

• Coefficient of volume expansion of glycerine, γ = 49 × 10^-5 °C^-1
• Rise in temperature, Δt = 30°C

When a liquid is heated its volume increases and hence its density decreases. If ρ and ρ′ are the densities at temperatures T and (T + ΔT) respectively, then

$\rho' = \rho(1 - \gamma\Delta \text t)$

Therefore the change in density is

$\rho - \rho' = \rho\gamma\Delta \text t$

and the fractional change in density is

$\dfrac{\rho - \rho'}{\rho} = \gamma\Delta \text t$

Substituting the values,

$\dfrac{\Delta \rho}{\rho} = (49 \times 10^{-5}) \times 30 \\[1em] = 1.47 \times 10^{-2}$

Hence, the fractional change in the density of glycerine is 1.47 × 10^-2, the density being decreased.

A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in the temperature of the block in 2.5 minutes, assuming that 50% of power is used up in heating the machine itself or lost to the surroundings. The specific heat of aluminium is 0.91 J g^-1 °C^-1.

Answer

Given,

• Power of the drilling machine, P = 10 kW = 10⁴ W
• Mass of the aluminium block, m = 8.0 kg
• Time, t = 2.5 min = 150 s
• Specific heat of aluminium, c = 0.91 J g^-1 °C^-1 = 910 J kg^-1 °C^-1
• Only 50% of the power is used in heating the block

The useful power supplied to the block is

$\text P' = \dfrac{50}{100} \times 10^4 = 5 \times 10^3\ \text W$

The heat given to the block in 150 s is

$\text Q = \text P' \times \text t = (5 \times 10^3) \times 150 \\[1em] = 7.5 \times 10^5\ \text J$

Using Q = m c ΔT, the rise in temperature is

$\Delta \text T = \dfrac{\text Q}{\text{mc}} = \dfrac{7.5 \times 10^5}{8.0 \times 910} \\[1em] = \dfrac{7.5 \times 10^5}{7280} = 103.0^\circ \text C$

Hence, the rise in the temperature of the aluminium block is about 103°C.

A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g^-1 °C^-1, heat of fusion of water = 335 J g^-1)

Answer

Given,

• Mass of the copper block, m = 2.5 kg
• Fall in temperature of the block, ΔT = 500 − 0 = 500°C
• Specific heat of copper, c = 0.39 J g^-1 °C^-1 = 390 J kg^-1 °C^-1
• Latent heat of fusion of ice, L = 335 J g^-1 = 3.35 × 10⁵ J kg^-1

The maximum amount of ice melts when the copper block cools right down to 0°C. The heat given out by the copper block is

$\text Q = \text{mc}\Delta \text T = 2.5 \times 390 \times 500 \\[1em] = 4.875 \times 10^5\ \text J$

If m′ be the mass of ice melted, then by the principle of calorimetry the heat lost by the copper block is used up as the latent heat of fusion of the ice,

$\text Q = \text m'\text L$

$\text m' = \dfrac{\text Q}{\text L} = \dfrac{4.875 \times 10^5}{3.35 \times 10^5} \\[1em] = 1.455\ \text{kg}$

Hence, the maximum amount of ice that can melt is about 1.5 kg.

In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm³ of water at 27°C. The final temperature is 40°C. Compute specific heat of the metal. The density of water is 10³ kg/m³ and the specific heat is 4.2 × 10³ J/(kg-°C).

If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value of specific heat of the metal?

Answer

Given,

• Mass of the metal block, m = 0.20 kg at 150°C
• Water equivalent of the calorimeter, W = 0.025 kg
• Volume of water = 150 cm³, density of water = 10³ kg m^-3, so mass of water m₁ = 0.15 kg at 27°C
• Final temperature of the mixture = 40°C
• Specific heat of water, c₁ = 4.2 × 10³ J kg^-1 °C^-1

Let c be the specific heat of the metal.

The heat lost by the metal block in cooling from 150°C to 40°C is

$\text Q_1 = \text{mc}(150 - 40) = 0.20 \times \text c \times 110 = 22\text c$

The heat gained by the water and the calorimeter in warming from 27°C to 40°C is

$\text Q_2 = (\text m_1 + \text W)\text c_1(40 - 27) \\[1em] = (0.15 + 0.025) \times (4.2 \times 10^3) \times 13 \\[1em] = 0.175 \times 4200 \times 13 = 9555\ \text J$

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

$22\text c = 9555$

$\text c = \dfrac{9555}{22} = 434.3$ J kg^-1°C^-1

Hence, the specific heat of the metal is about 0.43 × 10³ J kg^-1 °C^-1.

If the heat losses to the surroundings are not negligible, a part of the heat given out by the metal block escapes to the surroundings and does not reach the water and the calorimeter. The heat gained by the water and the calorimeter is then less than the heat actually lost by the block, and so the calculated value of c comes out to be less than the true value. Hence, the value obtained above is smaller than the actual specific heat of the metal.

Given below are observations on molar specific heats at room temperature of some common gases.

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K^-1. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

Answer

All the gases listed in the table are diatomic gases, whereas the value 2.92 cal mol^-1 K^-1 quoted for comparison belongs to a monatomic gas.

A monatomic gas molecule consists of a single atom, so it can possess only translational kinetic energy. The whole of the heat given to it at constant volume is used in increasing this translational energy, and hence its molar specific heat is small.

A diatomic gas molecule, on the other hand, possesses rotational energy in addition to translational energy. When heat is given to a diatomic gas at constant volume, a part of it is used in increasing the rotational energy of the molecules as well. Therefore more heat is required to raise the temperature of one mole of a diatomic gas through 1 K, and its molar specific heat is nearly 5 cal mol^-1 K^-1, which is markedly larger than that of a monatomic gas.

The value for chlorine (6.17 cal mol^-1 K^-1) is larger than that of the remaining diatomic gases. This indicates that in a chlorine molecule, besides the translational and the rotational motions, the vibrational motion of the two atoms about their mean positions is also appreciably excited at room temperature. The additional energy stored in this vibrational mode raises the molar specific heat of chlorine above the value for the other diatomic gases.

A 30 kg child running a temperature of 101°F is given antipyrin to increase the rate of evaporation of sweat from the body. As a result, the fever is brought down to 98°F in 20 minutes. Find the average rate of extra evaporation caused by the drug. The specific heat of human body is 4.18 × 10³ J kg^-1°C^-1 (same as of water) and the latent heat of evaporation of water at body temperature is 580 cal g^-1.

Answer

Given,

• Mass of the child, m = 30 kg
• Fall in temperature = 101°F − 98°F = 3°F
• Time, t = 20 minutes
• Specific heat of the human body, c = 4.18 × 10³ J kg^-1 °C^-1
• Latent heat of evaporation of water, L = 580 cal g^-1

A temperature difference of 1°F is equal to $\dfrac{5}{9}$°C. Hence the fall in temperature on the Celsius scale is

$\Delta \text T = 3 \times \dfrac{5}{9} = 1.667^\circ \text C$

The heat lost by the body of the child is

$\text Q = \text{mc}\Delta \text T = 30 \times (4.18 \times 10^3) \times 1.667 \\[1em] = 2.09 \times 10^5\ \text J$

Converting the latent heat of evaporation into joule per kilogram, using 1 cal = 4.2 J,

$\text L = 580 \times 4.2 = 2436\ \text{J g}^{-1} \\[1em] = 2.436 \times 10^6\ \text{J kg}^{-1}$

If m′ be the mass of sweat evaporated, then Q = m′L, so

$\text m' = \dfrac{\text Q}{\text L} = \dfrac{2.09 \times 10^5}{2.436 \times 10^6} \\[1em] = 0.0858\ \text{kg} = 85.8\ \text g$

The average rate of extra evaporation is

$\dfrac{\text m'}{\text t} = \dfrac{85.8}{20} = 4.29\ \text{g min}^{-1}$

Hence, the average rate of extra evaporation caused by the drug is about 4.3 g min^-1.

A 'thermacole' icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45°C, and coefficient of thermal conductivity of thermacole is 0.01 J s^-1 m^-1 K^-1. [Heat of fusion of water = 335 × 10³ J kg^-1].

Answer

Given,

• Side of the cubical icebox, a = 30 cm = 0.30 m
• Thickness of the wall, l = 5.0 cm = 0.05 m
• Mass of ice put in the box, m = 4.0 kg
• Time, t = 6 h = 6 × 3600 = 21600 s
• Temperature difference, θ₁ − θ₂ = 45 − 0 = 45°C
• Coefficient of thermal conductivity of thermacole, K = 0.01 J s^-1 m^-1 K^-1
• Latent heat of fusion of ice, L = 335 × 10³ J kg^-1

The box has six faces, so the total surface area through which heat enters is

$\text A = 6\text a^2 = 6 \times (0.30)^2 = 0.54\ \text m^2$

The quantity of heat entering the box in time t is

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

Substituting the values,

$\text Q = \dfrac{0.01 \times 0.54 \times 45 \times 21600}{0.05} \\[1em] = \dfrac{5248.8}{0.05} = 1.05 \times 10^5\ \text J$

The mass of ice melted by this heat is

$\text m' = \dfrac{\text Q}{\text L} = \dfrac{1.05 \times 10^5}{335 \times 10^3} = 0.313\ \text{kg}$

Therefore the mass of ice remaining after 6 hours is

$4.0 - 0.313 = 3.687\ \text{kg}$

Hence, about 3.7 kg of ice remains in the box after 6 hours.

A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s^-1 m^-1 °C^-1, heat of vaporisation of water = 2256 × 10³ J kg^-1.

Answer

Given,

• Base area of the boiler, A = 0.15 m²
• Thickness of the base, l = 1.0 cm = 1.0 × 10^-2 m
• Rate of boiling of water = 6.0 kg min^-1
• Thermal conductivity of brass, K = 109 J s^-1 m^-1 °C^-1
• Latent heat of vaporisation of water, L = 2256 × 10³ J kg^-1
• Temperature of the water = 100°C

The heat required per second to convert the water into steam is

$\dfrac{\text Q}{\text t} = \dfrac{6.0 \times (2256 \times 10^3)}{60} \\[1em] = 2.256 \times 10^5\ \text{J s}^{-1}$

Let θ be the temperature of the flame in contact with the boiler. In the steady state, this heat is conducted through the base of the boiler,

$\dfrac{\text Q}{\text t} = \dfrac{\text{KA}(\theta - 100)}{\text l}$

Substituting the values,

$2.256 \times 10^5 = \dfrac{109 \times 0.15 \times (\theta - 100)}{1.0 \times 10^{-2}}$

$2.256 \times 10^5 = 1635 \times (\theta - 100)$

$\theta - 100 = \dfrac{2.256 \times 10^5}{1635} = 138$

$\theta = 100 + 138 = 238^\circ \text C$

Hence, the temperature of the part of the flame in contact with the boiler is about 238°C.

Explain why:

(a) a body with large reflectivity is a poor emitter

(b) a brass tumbler feels much colder than a wooden tray on a chilly day

(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red-hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace

(d) the earth without its atmosphere would be inhospitably cold

(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.

Answer

(a) A body with large reflectivity reflects back most of the radiant energy falling on it, so only a small part of the incident energy is absorbed by it, that is, its absorptive power is small. According to Kirchhoff's law, at a given temperature the ratio of the emissive power to the absorptive power is the same for all bodies. Hence a body of small absorptive power must also have a small emissive power, that is, a body with large reflectivity is a poor emitter.

(b) Brass is a good conductor of heat while wood is a bad conductor. On a chilly day both the brass tumbler and the wooden tray are at the same low temperature. When we touch the brass tumbler, it conducts heat away from our hand very rapidly, and so it feels much colder. The wooden tray conducts heat away only very slowly, and so it does not feel so cold.

(c) An optical pyrometer is calibrated for the radiation of an ideal black body, for which the radiant energy emitted per second per unit area is σT⁴. A red-hot iron piece in the open is not a perfectly black body, so at the same temperature it emits less energy than a black body, that is, e σT⁴ where e is less than 1. The pyrometer therefore records too low a value of the temperature.

When the same piece is placed inside a furnace, it is surrounded by the walls of the furnace which are at the same temperature. The radiation coming out of the furnace is the radiation of a uniformly heated enclosure, which is very nearly black body radiation. Hence the pyrometer then gives the correct value of the temperature.

(d) The atmosphere is largely transparent to the incoming short-wavelength solar radiation, but it is partly opaque to the long-wavelength infra-red radiation emitted by the warm earth. The gases of the atmosphere absorb this infra-red radiation and re-emit a part of it back towards the surface, which keeps the surface warm. This is the greenhouse effect. In the absence of the atmosphere the whole of the infra-red radiation emitted by the earth would escape into space, and the earth would become inhospitably cold.

(e) Steam at 100°C carries with it the latent heat of vaporisation, which is 22.6 × 10⁵ J kg^-1, over and above the heat carried by water at 100°C. When steam condenses in the radiators of the heating system, it gives out this large amount of latent heat in addition to the heat given out on cooling. Hence, for the same mass of the circulating fluid, a steam heating system supplies much more heat than a hot water system, and is therefore more efficient in warming a building.

A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surrounding is 20°C.

Answer

Given,

• First case : the body cools from θ₁ = 80°C to θ₂ = 50°C in t = 5 minutes
• Second case : the body cools from θ₁ = 60°C to θ₂ = 30°C in time t′
• Temperature of the surroundings, θ₀ = 20°C

By Newton's law of cooling, the rate of loss of heat is directly proportional to the temperature difference between the body and its surroundings. Hence

$\dfrac{\theta_1 - \theta_2}{\text t} = \text K\left[\dfrac{\theta_1 + \theta_2}{2} - \theta_0\right]$

First case :

$\dfrac{80 - 50}{5} = \text K\left[\dfrac{80 + 50}{2} - 20\right] \\[1em] 6 = \text K(65 - 20) = 45\text K$

$\text K = \dfrac{6}{45} = \dfrac{2}{15}$

Second case :

$\dfrac{60 - 30}{\text t'} = \text K\left[\dfrac{60 + 30}{2} - 20\right] \\[1em] \dfrac{30}{\text t'} = \text K(45 - 20) = 25\text K$

Substituting the value of K,

$\dfrac{30}{\text t'} = 25 \times \dfrac{2}{15} = \dfrac{10}{3}$

$\text t' = \dfrac{30 \times 3}{10} = 9\ \text{minutes}$

Hence, the body takes 9 minutes to cool from 60°C to 30°C.