Calorimetry

Question

Two litres of water (density = 1 g/ml) in an open lid insulated kettle is heated by an electric heater of power 1 kW. The heat is lost from the lid at the rate of 160 J/s. The time taken for heating water (specific heat capacity $4.2\mathrm{kJ}{\mathrm{kg}}^{-1}{\mathrm{K}}^{-1}$) from 20°C to 75°C is ___seconds.

Moderate

Solution

Mass of 2 litres of water = 2 kg. Heat energy needed to raise the temperature of2 kg of water from 20°C to 75°C is

$Q=2\times \left(4.2\times {10}^{3}\right)\times 55=4.62\times {10}^{5}\mathrm{J}$

If t is the time taken, heat energy supplied by the heater in time t is

${\mathrm{Q}}_{1}=(\text{power}\times \text{time})=1000\mathrm{t}\text{joule}$

Heat energy lost in time t is

${\mathrm{Q}}_{2}=160\mathrm{t}\text{joule}$

Heat energy available for heating water is

$\mathrm{Q}={\mathrm{Q}}_{1}-{\mathrm{Q}}_{2}=840\mathrm{t}$

Equating $\mathrm{Q}={\mathrm{Q}}^{\mathrm{\prime}}$, we get $\mathrm{t}\simeq 550\mathrm{s}$.

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