Thermal capacity of \(40\text{ g}\) of aluminium of specific heat \(0.2\text{ cal/(g }^\{circ}\text{C)}\) is
Thermal capacity is given by the formula \(C = m \cdot s\). Substituting the values, \(C = 40\text{ g} \times 0.2\text{ cal/(g }^\circ\text{C)} = 8\text{ cal/}^\circ\text{C}\.
\(15\text{ gm}\) of ice at \(0^\circ\text{C}\) is mixed with \(300\text{ gm}\) of water at \(50^\circ\text{C}\) in a container. There is no heat loss due to radiation and water equivalent of container is ignored. What will be final temperature of water?
Heat absorbed to melt ice: \(Q_1 = 15 \times 80 = 1200\text{ cal}\). Let final temperature be \(T\). Heat gained by melted ice: \(15 T\). Heat lost by hot water: \(300(50 - T)\). Equilibrium: \(1200 + 15T = 300(50-T) \implies 315T = 13800 \implies T \approx 43.8^\circ\text{C}\).