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° C to 60° C is _________ s.
(Assume that the specific heat capacity of water remains constant over the temperature range of the water).

Solution

Physics sir · 3 steps

Only 70% of the heater's power reaches the water, so divide the heat needed by 1400 W, not 2000 W.

1. Heat the water needs. The temperature rises by 60 - 10 = 50 K. So Q = mcΔ T = 2 × 4200 × 50 = 4.2 × 10⁵ J.
2. Useful power. An efficiency of 70% means only 0.7 × 2000 = 1400 W actually goes into the water. The rest is lost to the surroundings.

   t = (mcΔ T)/(η P) = (4.2 × 10⁵ J)/(1400 W)

3. So. t = 300 s, so the answer is 300.

The common slip: 210. Ignoring the efficiency and dividing by the full 2000 W gives 210 s, which is too short.

The idea to keep. Efficiency always cuts the useful power. Heat needed equals efficiency times power times time, so a less efficient heater takes longer, never less.

Answer from the NTA final answer key. Solution by PrepWiser.