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 the heater is 70. Time required to heat 2 kg of water from 10^ C to…

Given: - Mass of water, m = 2 kg - Specific heat capacity, C = 4200 J kg^-1 K^-1 - Initial temperature, 10^ C - Final temperature, 60^ C - Heater power, P = 2000 W - Efficiency, 70 Find: Time required to heat the water. The energy required to heat the water is given by: Q = m C T where T = 60 - 10 = 50 K So, Q = 2 4200 50 = 420000 J The effective power of the heater is: P effective = 0.7 2000 = 1400 W The time required is: t = QP effective = 4200001400 = 300 s Therefore, the required time is 300 seconds.

A water heater of power 2000 W is used to heat water. The specifi… — Solution | JEE Physics

NVAEasyJEE 2023Calorimetry & Change of State

A water heater of power $2000 \, \text{W}$ is used to heat water. The specific heat capacity of water is $4200 \, \text{J kg}^{-1} \, \text{K}^{-1}$ . The efficiency of the heater is $70\%$ . Time required to heat $2 \, \text{kg}$ of water from $10^\circ \text{C}$ to $60^\circ \text{C}$ is ........ $\text{s}$ .

Answer

Correct answer:300

Step-by-step solution

Standard Method

Given:

Mass of water, $m = 2 \, \text{kg}$
Specific heat capacity, $C = 4200 \, \text{J kg}^{-1} \, \text{K}^{-1}$
Initial temperature, $10^\circ \text{C}$
Final temperature, $60^\circ \text{C}$
Heater power, $P = 2000 \, \text{W}$
Efficiency, $70\%$

Find: Time required to heat the water.

The energy required to heat the water is given by:

Q = m C \Delta T

where

\Delta T = 60 - 10 = 50 \, \text{K}

So,

Q = 2 \times 4200 \times 50 = 420000 \, \text{J}

The effective power of the heater is:

P_{\text{effective}} = 0.7 \times 2000 = 1400 \, \text{W}

The time required is:

t = \frac{Q}{P_{\text{effective}}} = \frac{420000}{1400} = 300 \, \text{s}

Therefore, the required time is $300$ seconds.

Common mistakes

Using $2000 \, \text{W}$ directly as the useful power is incorrect because the heater efficiency is only $70\%$ . First calculate the effective power using $0.7 \times 2000$ .
Taking temperature change as $60$ or $10$ instead of $60 - 10 = 50 \, \text{K}$ is wrong. Always use the difference in temperature for $\Delta T$ .
Confusing degree Celsius difference with an issue in units is unnecessary here. A change of $50^\circ \text{C}$ is numerically the same as $50 \, \text{K}$ , so use $\Delta T = 50$ in the formula.

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