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JEE Advanced · Physics · 12. Thermal Properties

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

  1. A 2
  2. B 4
  3. C 6
  4. D 8
Verified Solution

Answer & Solution

Correct Answer

(D) 8

Step-by-step Solution

Detailed explanation

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 .
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