Question

Take To - 20°C In order to cool 1 ton of water at 20°C in an insulated tank, a person pours 64 kg of ice at -8°C into the water. Given the melting temperature and the heat of fusion of ice at atmospheric pressure are 0°C and 333.7 kJ/kg, respectively, determine (a) the final equilibrium temperature in the tank (Round to two decimal places) and (b) the exergy destroyed during this process. X destroyed kJ (Round to the nearest integer) View Details of Last Check Answer C

Answer 1

Mass of water , M_w = stun = 1000 kg mass of ice , m_i = 64 kg At Equilibrium, Heat Gain by ice = Heat release by water implies (Heat absorbed in solid phase + Latent Heat + Heat absorbed ln Liquid phase) = Heat released by water implies (C_ice = S_p . Heat of ice = 2.093 KJ/kg) (C_pau = 4.187 KJ/kg) implies m_i C_ice (273 - (265)] + m_iL + m_i C_w (T_1 - 273) = m_w C_ice times (293 - T_1) implies 64 times 2.093 times (8) + 64 times (3337) + 64 times 4.187 (T_1 - 273) = 1000 times 4.187 times (293 - T_1) implies -50726.84 + 26 7.96 T_1 = 122679.1 - 4187 T_1 implies T_1 = 286.76 K Therefore T_1 in degree C = 286.76 - 273 = 13.763 degree C T_1 = 13.763 degree C Answer b) Energy destroyed = Generation of Entropy = (unviable Energy)