Q 11-10-082JEE MainJEE Main 2022 (25 Jul, Shift 2)Easy
A block of ice of mass $120\ \text{g}$ at temperature $0^\circ\text{C}$ is put in $300\ \text{g}$ of water at $25^\circ\text{C}$. $x$ g of ice melts as the temperature of the water reaches $0^\circ\text{C}$. The value of $x$ is ______. [Use: specific heat capacity of water $= 4200\ \text{J kg}^{-1}\text{K}^{-1}$, latent heat of ice $= 3.5\times10^5\ \text{J kg}^{-1}$]
Numerical value type. Enter your answer.
Answer: 90
Heat released by the water cooling to $0^\circ\text{C}$: $0.3\times4200\times25 = 31500\ \text{J}$.
$$m = \frac{31500}{3.5\times10^5} = 0.09\ \text{kg} = 90\ \text{g}$$
Solution by Sreeraj P, M.Sc Physics