Question

Two litres of water kept in a container at \( 27^\circ \text{C} \) is heated with a coil of 1 kW. The lid of the container is open and energy dissipates at the rate of 160 J/s. If the specific heat of water is 4.2 kJ/kg, then the time taken by coil to raise the temperature of water from \( 27^\circ \text{C} \) to \( 77^\circ \text{C} \) is:

• A. 840 s
• B. 500 s
• C. 420 s
• D. 372 s

Hint:

When heating water, take into account both the power input and energy lost due to dissipation. The effective power determines the heating time.

Answer 1

The Correct Option is B

The heat required to raise the temperature of the water is given by the formula: \[ Q = mc\Delta T \] where: - \( m = 2 \, \text{kg} \) (since 2 litres of water is 2 kg) - \( c = 4200 \, \text{J/kg}^\circ \text{C} \) (specific heat of water) - \( \Delta T = 77^\circ \text{C} - 27^\circ \text{C} = 50^\circ \text{C} \) Thus, the heat required is: \[ Q = 2 \times 4200 \times 50 = 420000 \, \text{J} \] The power supplied by the coil is 1 kW, i.e., \( 1000 \, \text{J/s} \).
However, the energy is dissipating at a rate of 160 J/s, so the effective power heating the water is: \[ P_{\text{effective}} = 1000 - 160 = 840 \, \text{J/s} \] Now, the time required to heat the water is: \[ t = \frac{Q}{P_{\text{effective}}} = \frac{420000}{840} = 500 \, \text{s} \]
Thus, the correct answer is (b).