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.

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 $\mathrm{t}\simeq 550\mathrm{s}$.