Q 11-10-065JEE MainJEE Main 2023 (31 Jan, Shift 2)Easy
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 heater is $70\%$. Time required to heat $2\ \text{kg}$ of water from $10^\circ\text{C}$ to $60^\circ\text{C}$ is ______ s. (Assume that the specific heat capacity of water remains constant over the temperature range of the water.)
Numerical value type. Enter your answer.
Answer: 300
Heat needed $=2\times4200\times50=4.2\times10^5\ \text{J}$. Useful power $=0.7\times2000=1400\ \text{W}$.
$t=\dfrac{4.2\times10^5}{1400}=300\ \text{s}$.
Solution by Sreeraj P, M.Sc Physics