The two metals A and B have their specific heat capacities in the ratio 2:3. If they are supplied the same amount of heat, then

(a) which metal piece will show a greater rise in temperature if their masses are the same?

(b) calculate the ratio in which their temperatures rise, if the mass ratio of metal A and metal B is 3:5.

The rise in temperature is given by: ΔT = $\dfrac{\text{Q}}{\text{mc}}$
Where:

• Q = heat supplied (same for both)
• m = mass
• c = specific heat capacity
• Δ T = temperature change

(a) Given that masses are same,

Since: $ΔT = \dfrac{\text{Q}}{\text{m c}}$
here Q and m are the same for both metals A and B, so $ΔT \propto \dfrac{1}{\text{c}}$

So, $\dfrac{ΔT$A} {ΔTB} = \dfrac{cB}{cA} =\dfrac{3}{2} = 3 : 2
Metal A shows a greater temperature rise than metal B (since it has lower specific heat).

(b) Masses in ratio 3 : 5 Now :

$\dfrac{ΔT$A} {ΔTB} = \dfrac{mB \cdot cB}{mA\cdot cA} \\[1em] \dfrac{ΔTA} {ΔTB} = \dfrac{5 \cdot 3}{3 \cdot 2} \\[1em] \dfrac{ΔTA} {ΔTB} = \dfrac{15}{6} \\[1em] \dfrac{ΔTA} {ΔTB} = \dfrac{5}{2} \\[1em] \\[1em] {ΔTA} :{ΔTB} =5 : 2

The ratio of temperature rise is 5 : 2