10 kg of ice at $-10^\circ\text{C}$ is added to 100 kg of water to lower its temperature from 25 $^\circ\text{C}$. Consider no heat exchange to surroundings. The decrement to the temperature of water is ______ $^\circ\text{C}$.
(specific heat of ice $= 2100\ \text{J/Kg}.^\circ\text{C}$, specific heat of water $= 4200\ \text{J/Kg}.^\circ\text{C}$, latent heat of fusion of ice $= 3.36\times10^5\ \text{J/Kg}$ )
Answer: (B) 10
Heat needed to bring the ice to water at $0^\circ$C:
$$10\times2100\times10 + 10\times3.36\times10^5 = 2.1\times10^5 + 33.6\times10^5 = 35.7\times10^5\ \text{J}$$
Heat the water can give by cooling to $0^\circ$C: $100\times4200\times25 = 105\times10^5$ J, which is more, so all the ice melts. Let the final temperature be $T$:
$$35.7\times10^5 + 10\times4200\,T = 100\times4200\,(25 - T)$$
$$35.7\times10^5 + 0.42\times10^5\,T = 105\times10^5 - 4.2\times10^5\,T$$
$$4.62\times10^5\,T = 69.3\times10^5 \Rightarrow T = 15^\circ\text{C}$$
Decrease in the water's temperature $= 25 - 15 = 10^\circ$C.
Solution by Sreeraj P, M.Sc Physics