There is 0.2 Kg of water at 25^(0)C in a calorimeter of heat capacity 62.8 J K^(-1). If the temperature of the calorimeter is to be raised to 50^(0)C, how much steam at 100°C is to be sent into it? (L_("steam") = 2.26 xx 10^(6)JKg^(-1), specific heat of water = 4186J Kg^(-1))

There is 0.2 Kg of water at 25^(0)C in a calorimeter of heat capacity

1.	There is 0.2 Kg of water at 25^(0)C in a calorimeter of heat capacity 62.8 J K^(-1). If the temperature of the calorimeter is to be raised to 50^(0)C, how much steam at 100°C is to be sent into it? (L_("steam") = 2.26 xx 10^(6)JKg^(-1), specific heat of water = 4186J Kg^(-1))
Answer» Solution :Heat capacity of the calorimeter (mass x SPECIFIC heat) `= 62.8 J K^(-1)` Mass of water `(m_(2)) = 0.2 KG` , Specific heat of water `(S_(2)) = 4186 J Kg^(-1)` Initial temperature `(t_(1)) = 25^(0)C` , Final temperature `(t_(2)) = 50^(0)C` Temperature of the steam `(t_(3)) = 100^(0)C` Heat gained by the calorimeter `= m_(1) s_(1) (t_(2) - t_(1)) = 62.8 xx 25 = 1570 J` Heat gained by the water `= m_(2)S_(2) (t_(2)-t_(1)) = 0.2 xx 4186 xx 25 = 20930 J` Heat lost by the steam `= m_(3) L + m_(3)S_(2)(t_(3)-t_(2)) = m_(3) [L + S_(2) (t_(3) - t_(2))]` `= m_(3) (2.26 xx 10^(6) + 4186 xx (100 - 50)] = m_(3) [2260000 + 209300]` `= m_(3) xx 2469300 J.` From the method of mixtures Heat lost by steam =Heat gained by the calorimeter + heat gained by the water `m_(3) xx 2469300 = 1570+ 20930= 22500 m_(3) = (22500)/(2469300)=0.009112 Kg` `therefore` Mass of steam required = 0.009112 Kg.