1. The problem statement, all variables and given/known data Cooling of a food from 293K (20 ⁰C), freezing it at 272K (-1⁰C), and further cooling to 253K (-20 ⁰C); using cooling water rising in temperature from 293K (20 ⁰C) to 303K (30 ⁰C) to accept heat. The specific heat of the unfrozen food is taken as 3.2 kJ/kg K, that of the frozen food as 1.7 kJ/kg K, and the enthalpy of the freezing as 240 kJ/kg. Calculate the Lorenz coefficient of performance and compare it with Carnot coefficient of performance. 2. Relevant equations 3. The attempt at a solution Lorenz COP: For cooling water side, the average absolute temperature is simply the logarithmic mean absolute temperature: T2,avg = (303-293)/ln(303-293) = 298 K For the food, the entropy change can be calculated as follows: s1'- s1'' = (1.7 ln (272/253)+240/272+3.2 ln (293/272))=1.243 kJ/kgK The heat per unit mass is: q1=1.7×(272-253)+240+3.2×(293-272)=339.5 kJ/kg The average absolute temperature of acceptance of heat: T1,avg=339.5/1.243 = 273.1 K The Lorenz COP is thus: COP= (273.1)/(298-273.1)=10.97 For the Reversed Carnot cycle, it has to operate between the highest and lowest temperatures to satisfy the requirement for heat transfer. This leads to: COP= (253)/(303-253)=5.06 Are the calculations correct? I tried to confirm the answer with my teacher, who told me that the Carnot COP is incorrect. He says the correct answer is: Carnot COP = Tlow/(Thigh - Tlow) COP= (293)/(293-293) = ∞ He told me that in Carnot refrigeration, the temperature of the food and the cooling water remain constant (unchanging) and thus we should use the initial temperatures for both the food (293K before cooling) and the water (293K before receiving any heat). Is he correct? I couldn't seem to really understand the reason behind his answer. Could someone please kindly verify the answers? Thank you for the help.