An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0° C to 15.0° C in 100 s. Calculate — (i) the heat capacity of 4.0 kg of liquid, and (ii) the specific heat capacity of liquid.

(i) Power of heater (P) = 600 W

Mass of liquid (m) = 4.0 kg

Change in temperature of liquid = (15 – 10)°C = 5° C (or 5 K)

Time taken to raise it's temperature (t) = 100 s

heat capacity = ?

From relation,

$Q = \text {Power} \times \text{time}$

Substituting the values in the formula above we get,

$Q = 600 \times 100 \\[0.5em] \Rightarrow Q = 60000 J \\[0.5em]$

Now,

$C^{'} = \dfrac{Q}{△T} \\[0.5em]$

Substituting the values in the formula above we get,

$C^{'} = \dfrac{60,000}{5} \\[0.5em] C^{'} = 1.2 \times 10^{4} \text{ J} \text{K}^{-1} \\[0.5em]$

Hence, heat capacity = 1.2 x 10 J K^-1

(ii) specific heat capacity {c} = ?

$c = \dfrac{Q}{m \times △T} \\[0.5em]$

Substituting the values in the formula above we get,

$c = \dfrac{60000}{4 \times 5} \\[0.5em] c = \dfrac{60000}{4 \times 5} \\[0.5em] c = 3 \times 10^{3} \text{ J kg}^{-1} \text{K}^{-1}$

Hence, specific heat capacity = 3 x 10³ J Kg^-1 K^-1