An iron piece of mass 200 ~g is kept inside a furnace for some time and then put in a calorimeter of water equivalent 20 ~g containing 230 ~g of water at 20 C . The steady state temperature attained by the mixture is 60^ . The temperature of the furnace is (Specific heat capacity of iron is 470 ~J~kg ⁻¹ C ⁻¹ )

Step-by-step solution

Let T be the temperature of the furnace. The heat lost by the iron piece is equal to the heat gained by the calorimeter and the water. Mass of iron m_ Fe = 0.2 kg . Specific heat of iron c_ Fe = 470 J kg ⁻¹ C ⁻¹ . Water equivalent of calorimeter w = 0.02 kg . Mass of water m_w = 0.23 kg . Specific heat of water c_w = 4200 J kg ⁻¹ C ⁻¹ . Initial temperature of water and calorimeter T_i = 20^ C . Final steady state temperature T_f = 60^ C . Heat lost by iron = m_ Fe c_ Fe (T - T_f) = 0.2 470 (T - 60) = 94(T - 60) . H