An iron piece of mass 100 g is placed inside a furnace for a long duration and subsequently dropped into a calorimeter of water equivalent 10 g that contains 240 g of water at 20^ C . The mixture reaches a final equilibrium temperature of 60^ C . Determine the temperature of the furnace. The specific heat capacity of iron is 470 J kg ⁻¹ ^ C ⁻¹ .

Step-by-step solution

Let the temperature of the furnace be T . Mass of the iron piece, m_i = 100 g = 0.1 kg Specific heat capacity of iron, c_i = 470 J kg ⁻¹ ^ C ⁻¹ Total effective mass of water (water + calorimeter), M = 240 g + 10 g = 250 g = 0.25 kg Specific heat capacity of water, c_w = 4200 J kg ⁻¹ ^ C ⁻¹ According to the principle of calorimetry, heat lost by the iron piece equals the heat gained by the water and the calorimeter. m_i c_i (T - T_f) = M c_w (T_f - T_i) Substituting the given values: 0.1 470 (T - 60) = 0.25 4200 (60