Question

An electric heater of power 1 kW raises the temperature of 500 g of a given liquid from 25°C to 31°C in 2 minutes. Calculate:

1. Heat capacity of the liquid.
2. Specific heat capacity of the liquid.

Answer 1

Given: Power of electric heater,

\[P = 1\ \text{kW} = 1000\ \text{W}\]

Mass of liquid,

\[m = 500\ \text{g} = 0.5\ \text{kg}\]

Initial temperature,

\[T_{1} = 25^{\circ}\text{C}\]

Final temperature,

\[T_{2} = 31^{\circ}\text{C}\]

Time,

\[t = 2\ \text{min} = 120\ \text{s}\]

Let the required heat capacity be \[C\]

(i) Heat capacity of the liquid.

Formula: Energy supplied by the heater, using the formula,

\[Q = P t\]

Substituting the values,

\[Q = 1000 \times 120\]

\[= 120000\ \text{J}\]

Rise in temperature, using the formula,

\[\Delta T = T_{2} - T_{1}\]

Substituting the values,

\[\Delta T = 31 - 25\]

\[= 6^{\circ}\text{C}\]

Heat capacity of the liquid, using the formula,

\[C = \frac{Q}{\Delta T}\]

Substituting the values,

\[C = \frac{120000}{6}\]

\[= 20000\ \text{J}/^{\circ}\text{C}\]

Hence, the required heat capacity of the liquid is \[20000\ \text{J}/^{\circ}\text{C}\]

(ii) Specific heat capacity of the liquid.

Let the required specific heat capacity be \[c\]

Specific heat capacity of the liquid, using the formula,

\[c = \frac{Q}{m \Delta T}\]

Substituting the values,

\[c = \frac{120000}{0.5 \times 6}\]

\[= \frac{120000}{3}\]

\[= 40000\ \text{J kg}^{-1}\,^{\circ}\text{C}^{-1}\]

Hence, the required specific heat capacity of the liquid is \[40000\ \text{J kg}^{-1}\,^{\circ}\text{C}^{-1}\]