A piece of metal of 850 K is dropped into 1 kg of water at 300 K. If equilibrium temperature of water is 350 K then the heat capacity of the metal expressed in J K is ( Sp. heat of water = 1 cal g ^ C )

Step-by-step solution

According to the principle of calorimetry, heat lost by the metal is equal to the heat gained by the water. Heat gained by water is Q = m_w s_w T_w Given mass of water m_w = 1 kg = 1000 g Specific heat of water s_w = 1 cal/g ^ C = 4.2 J/g ^ C Change in temperature of water T_w = 350 - 300 = 50 K Q = 1000 4.2 50 = 210000 J Heat lost by the metal is Q = C T_m Change in temperature of the metal T_m = 850 - 350 = 500 K Equating the heat lost to the heat gained: C 500 = 210000 C = 210000 500 = 420 J/K Answer: 420