Q 11-10-124JEE MainJEE Main 2017 (2 Apr)Easy
A copper ball of mass $100$ g is at a temperature $T$. It is dropped in a copper calorimeter of mass $100$ g, filled with $170$ g of water at room temperature. Subsequently, the temperature of the system is found to be $75^\circ$C. $T$ is given by:
(Given: room temperature $= 30^\circ$C, specific heat of copper $= 0.1\ \text{cal g}^{-1}\,{}^\circ\text{C}^{-1}$)
Answer: (C) $885^\circ$C
Heat lost by the ball = heat gained by the calorimeter and water (each rises from $30^\circ$C to $75^\circ$C, i.e. by $45^\circ$C):
$$100(0.1)(T - 75) = 100(0.1)(45) + 170(1)(45)$$
$$10(T-75) = 450 + 7650 = 8100 \;\Rightarrow\; T - 75 = 810$$
$$T = 885^\circ\text{C}$$
Solution by Sreeraj P, M.Sc Physics