Averages rate of cooling:
200 kg of ice at - 10oC is located in a bunker to cool some vegetables. 24 hours afterwards the ice has melted into water at 50C. Finds averages rate of cooling in kJ/hr & TR provided through the ice? Regard as
Specific heat of ice, cp,i = 1.94 kJ/kg oC
Specific heat of water, cp,w = 4.1868 kJ/kg oC
Latent heat of fusion of ice at 0oC, L = 335 kJ/kg.
Solution:
Rate of addition heat to 200 kg ice at 10oC to formulate water at 5oC
QC = (sensible heat gain in ice from - 10oC to 0oC + latent heat of melting of ice + sensible heat gain of water from 0oC to 5oC)/time
= (200 kg * (0 - (- 10)) oC * 1.94 kJ/kg. oC + 200 kg * 335 kJ/kg + 200 kg * (5 - 0) oC* 4.1868 kJ/kg. oC)/24 hr
= 3127.78 kJ/hr
