QJEEM23Ph0904: JEE Main 2023 Physics - Class 11 JEEM23Ph0904 Solution: A water heater of power is used to heat water. The specific…

The amount of heat energy required to raise the temperature of a substance can be calculated as:

Q = m $\times$ c $\times$ ΔT

where Q is the heat energy required, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.

The time required to heat a substance can be calculated as :

t = ${Q \over P}$

where t is the time required, and P is the power of the heating device.

The actual power output of the heating device can be calculated as:

P_actual = P_input $\times$ efficiency

where P_input is the input power to the device and efficiency is the fraction of input power that is actually converted to useful power output.

Substituting the given values:

Q = 2 kg $\times$ 4200 J/kg/K $\times$ (60 - 10) = 2 kg $\times$ 4200 J/kg/K $\times$ 50 K = 4200 $\times$ 50 $\times$ 2 J = 420,000 J

P_input = 2000 W = 2000 J/s

P_actual = 2000 $\times$ 0.7 = 1400 J/s

t = ${Q \over {{P_{actual}}}}$ = ${{420,000} \over {1400}}$ J/s = 300 s

So, the time required to heat 2 kg of water from 10°C to 60°C is approximately 300 s.