A metal ball of mass $0.1$ kg is heated up to $500^\circ$C and dropped into a vessel of heat capacity $800\ \text{J K}^{-1}$ and containing $0.5$ kg water. The initial temperature of water and vessel is $30^\circ$C. What is the approximate percentage increment in the temperature of the water? [Specific heat capacities of water and metal are, respectively, $4200\ \text{J kg}^{-1}\text{K}^{-1}$ and $400\ \text{J kg}^{-1}\text{K}^{-1}$]
Answer: (D) $20\%$
Heat lost by the ball = heat gained by water and vessel:
$$0.1\times400(500 - T) = (0.5\times4200 + 800)(T - 30)$$
$$40(500 - T) = 2900(T - 30) \Rightarrow 2940T = 107000 \Rightarrow T \approx 36.4^\circ\text{C}$$
Increase $\approx 6.4^\circ$C on $30^\circ$C, i.e. about $21\% \approx 20\%$.
Solution by Sreeraj P, M.Sc Physics