Question

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 $10^{\circ} C$ to $60^{\circ} C$ is___$ s$
(Assume that the specific heat capacity of water remains constant over the temperature range of the water)

Hint:

To find the time required to heat the water, use the effective power accounting for efficiency.

Answer 1

Correct Answer: 300

To solve the problem, we first determine the heat energy required to raise the temperature of the water. The formula to calculate the heat energy (Q) is:

Q = mcΔT

where:
m = mass of water = 2 kg,
c = specific heat capacity of water = 4200 J kg^-1 K^-1,
ΔT = change in temperature = (60 - 10)°C = 50 K.

Substituting the values, we get:

Q = 2 kg × 4200 J kg^-1 K^-1 × 50 K = 420,000 J

Since the heater has an efficiency of 70%, only 70% of the power is used effectively to heat the water. Therefore, the useful power (P_effective) is:

P_effective = 2000 W × 0.70 = 1400 W

Next, we calculate the time (t) required using the formula:

t = Q / P_effective

Substituting the values, we have:

t = 420,000 J / 1400 W ≈ 300 s

The calculated time is approximately 300 seconds, which fits within the expected range of (300, 300) seconds.