Question

A water heater of power $2000 W$ is used to heat water. The specific heat capacity of water is $4200 J$ $kg ^{-1} K ^{-1}$ .The efficiency of heater is $70 \%$ .Time required to heat $2 kg$ of water from $10^{\circ} C$ to $60^{\circ} C$ is___$ s$
(Assume that the specific heat capacity of water remains constant over the temperature range of the water)

Hint:

To find the time required to heat the water, use the effective power accounting for efficiency.

Answer 1

Correct Answer: 300

The energy required to heat the water is given by: \[ Q = m C \Delta T \] where \( m = 2 \, \text{kg} \), \( C = 4200 \, \text{J/kg K} \), and \( \Delta T = 60 - 10 = 50 \, \text{K} \). Thus, the total energy required: \[ Q = 2 \times 4200 \times 50 = 420000 \, \text{J} \] The power of the heater is 2000 W, but since the efficiency is 70\%, the effective power is: \[ P_{\text{effective}} = 0.7 \times 2000 = 1400 \, \text{W} \] The time required to heat the water is: \[ t = \frac{Q}{P_{\text{effective}}} = \frac{420000}{1400} = 300 \, \text{seconds} \]