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JEE Advanced 2010Paper 1
Exam
JEE Advanced
Year
2010
Paper
Paper 1
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Advanced 2010 · Paper 1 · Physics · Thermal Properties of Matter

A piece of ice (heat capacity =2100 ~J ~kg^ -1 ^ C^ -1 and latent...

A piece of ice (heat capacity =2100 J kg−1∘C−1=2100 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1} and latent heat =3.36×105 J kg−1=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1} ) of mass m\mathrm{m} gram is at −5∘C-5^{\circ} \mathrm{C} at atmospheric pressure. It is given 420 J420 \mathrm{~J} of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that 1 g1 \mathrm{~g} of ice has melted. Assuming there is no other heat exchange in the process, the value of mm is

Correct answer

0

Step-by-step solution

Language of question is slightly wrong. As heat capacity and specific heat are two different physical quantities. Unit J−kg−1⋅∘C−1\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. 420=(m×10−3)(2100)(5)+(1×10−3)(3.36×105)\begin{array}{r}420=\left(m \times10^{-3}\right)(2100)(5)+\left(1 \times10^{-3}\right) \\ \left(3.36 \times10^5\right) \end{array} Solving this equation we get, m=8 \mathrm{~g}\therefore The correct answer is 8 .
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