A 200-L electric water heater uses 2.0 kW. Assuming no heat loss, how many hours would it take to heat the water in this tank from 23°C to 75°C? The specific heat of water is 4186 J/kg ∙ K and its density is 1000 kg/m³.

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Quizplus · Accepted Answer

The energy required to heat the water is calculated using Q = mcΔT. The mass m is 200 kg (since density is 1000 kg/m³ and volume is 0.2 m³), c is 4186 J/kg∙K, and ΔT is 52 K (75°C - 23°C). The energy required is 200 kg * 4186 J/kg∙K * 52 K = 43,139,200 J. The power of the heater is 2.0 kW, which is 2000 J/s. Time is energy divided by power, so 43,139,200 J / 2000 J/s = 21,569.6 seconds, which is approximately 6 hours.

A 200-L Electric Water Heater Uses 2