A 200-L electric water heater uses 2.0 kW.Assuming no thermal energy loss,how many hours would it take to change the temperature of 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.

Question

Quizplus · Accepted Answer

The amount of thermal energy needed to heat up the water in the tank is:

Q = m * c * ΔT

where
m = mass of water = density * volume = 1000 kg/m³ * 0.2 m³ = 200 kg
c = specific heat of water = 4186 J/kg∙K 
ΔT = change in temperature = 75°C - 23°C = 52°C

So, Q = 200 kg * 4186 J/kg∙K * 52°C = 4,346,240 J

The time required to generate this amount of thermal energy at a rate of 2.0 kW is:

t = Q / P = 4,346,240 J / 2,000 W = 2173.12 s = 36.22 min = 0.604 h

Therefore, the time required to change the temperature of water in the tank from 23°C to 75°C is approximately 6.0 hours (rounded to the nearest hour).

So, the best choice is B.

A 200-L Electric Water Heater Uses 2