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.

Question

Quizplus · Accepted Answer

The amount of heat required to raise the temperature of 200 L of water from 23°C to 75°C can be calculated as follows:

Q = m × c × ΔT
where Q is the heat required (in Joules), m is the mass of water (in kg), c is the specific heat of water (in J/kg∙K), and ΔT is the temperature difference (in K).

m = density × volume
where density is 1000 kg/m³ and volume is 200 L = 0.2 m³

m = 1000 kg/m³ × 0.2 m³ = 200 kg

ΔT = 75°C - 23°C = 52°C

Q = 200 kg × 4186 J/kg∙K × 52 K = 4.32 × 10⁸ J

The power of the electric water heater is 2.0 kW = 2.0 × 10³ W.

The time taken to heat the water can be calculated as follows:

time = Q / power

time = 4.32 × 10⁸ J / 2.0 × 10³ W = 2.16 × 10⁵ s

time = 2.16 × 10⁵ s / 3600 s/h = 60 h

Therefore, the answer is B) 6.0 hours, assuming no heat loss.

A 200-L Electric Water Heater Uses 2