Solved Sample Paper 2 — Question 27

Given,

Amount of heat supplied (Q)= 2240 J

Rise in temperature (△t) = (65 – 25)°C = 40°C

Heat capacity = ?

(a) Heat capacity (C') = $\dfrac{\text{\text{Q}}}{\text{△t}}$

Substituting the values in the formula above we get,

$= \dfrac{2240}{40} = 56 \text{ J°C}^{-1}$

Hence, heat capacity of metal piece = 56 J per °C

(b) Specific heat capacity (c) = ?

mass (m) = 450 g

We know that,

$\text{c} = \dfrac{\text{C}'}{\text{m}}$

Substituting the values in the formula above we get,

$\text{c} = \dfrac{56}{450} = 0.12\text{ J g}^{-1} \text{°C}^{-1}$

Hence, Specific heat capacity = 0.12 J g^-1 °C^-1