Q 11-10-121JEE MainJEE Main 2019 (12 Apr, Shift 1)Easy
When $M_1$ gram of ice at $-10^\circ$C (specific heat $= 0.5\ \text{cal g}^{-1}{}^\circ\text{C}^{-1}$) is added to $M_2$ gram of water at $50^\circ$C, finally no ice is left and the water is at $0^\circ$C. The value of latent heat of ice, in cal g$^{-1}$ is:
Answer: (D) $\dfrac{50M_2}{M_1} - 5$
Heat gained by ice = heat lost by water:
$$M_1(0.5)(10) + M_1L = M_2(1)(50) \Rightarrow L = \frac{50M_2}{M_1} - 5$$
Solution by Sreeraj P, M.Sc Physics