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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)

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answer is 300.

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Detailed Solution

Given power = 2000W and efficiency = 70%

Heat required to increase the temperature of water by $∆ T$ is

Heat = $ms ∆ T$

Therefore,

$(\frac{70}{100} P = \frac{msΔT}{t})$

$\frac{70}{100} (2000) = \frac{2 \times 4200 \times 50}{t}$

$t = \frac{2 \times 4200 \times 50}{70 \times 20} = 300 \sec$

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