A piece of ice (heat capacity $=2100 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ and latent…

A piece of ice (heat capacity $=2100 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ and latent heat $=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$ ) of mass $\mathrm{m}$ gram is at $-5^{\circ} \mathrm{C}$ at atmospheric pressure. It is given $420 \mathrm{~J}$ of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that $1 \mathrm{~g}$ of ice has melted. Assuming there is no other heat exchange in the process, the value of $m$ is

Solution

Language of question is slightly wrong. As heat capacity and specific heat are two different physical quantities. Unit $\mathrm{J}-\mathrm{kg}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}$. The heat capacity given in the question is really the specific heat. Now applying the heat exchange equation. $ \begin{array}{r} 420=\left(m \times 10^{-3}\right)(2100)(5)+\left(1 \times 10^{-3}\right) \\ \left(3.36 \times 10^5\right) \end{array} $ Solving this equation we get, $ m=8 \mathrm{~g} $ $\therefore$ The correct answer is 8

Asked in: JEE Advanced 2010 (Paper 1)

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