A block of ice of mass 150 g at 0^ C is dropped into 250 g of water at T^ C . It is observed that exactly 100 g of ice melts by the time the mixture reaches thermal equilibrium. The value of T is ________. [Use: Specific heat capacity of water = 4200 J kg ⁻¹ K ⁻¹ , Latent heat of fusion of ice = 3.36 10^5 J kg ⁻¹ ]

Step-by-step solution

Since only a part of the ice melts ( 100 g out of 150 g ), the final temperature of the mixture must be 0^ C . According to the principle of calorimetry, the heat lost by the water equals the heat gained by the melting ice. Heat lost by water: Q₁ = m_w s_w T Q₁ = 0.25 4200 (T - 0) = 1050 T J Heat gained by ice to melt 100 g : Q₂ = m_ melt L Q₂ = 0.1 3.36 10^5 = 33600 J Equating the heat lost and heat gained: 1050 T = 33600 T = 33600 1050 = 32 The initial temperature of the water was 32^ C . Answer: 32