For purposes of these numerics, take the specific heat of ice as 2 kJ/kg K, the
ID: 1989606 • Letter: F
Question
For purposes of these numerics, take the specific heat of ice as 2 kJ/kg K, the heat of fusion for ice as 300 kJ/kg, and the specific heat of water as 4 kJ/kg K. How much heat energy (Qin ) is required (extracted or added) to take 100 g of water at 50 °C, freeze it, and lower the temperature of the ice to -100 °C? For purposes of these numerics, take the specific heat of ice as 2 kJ/kg K, the heat of fusion for ice as 300 kJ/kg, and the specific heat of water as 4 kJ/kg K. How much heat energy (Qin ) is required (extracted or added) to take 100 g of water at 50 °C, freeze it, and lower the temperature of the ice to -100 °C? For purposes of these numerics, take the specific heat of ice as 2 kJ/kg K, the heat of fusion for ice as 300 kJ/kg, and the specific heat of water as 4 kJ/kg K. How much heat energy (Qin ) is required (extracted or added) to take 100 g of water at 50 °C, freeze it, and lower the temperature of the ice to -100 °C?Explanation / Answer
HEat required,Qm=m*[sw*delta(T) +Hf+Sicedelta(T)] =0.1*[4*(50)+300+2*100] =70KJ
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