An autoclave contains 1000 cans of pea soup. It is heated to an overall temperature of 100°C. If the cans are to be cooled to 40°C before leaving the autoclave, how much cooling water is required if it enters at 15°C and leaves at 35°C? The specific heats of the pea soup and the can metal are respectively 4.1 kJ kg-1 °C-1 and 0.50 kJ kg-1 °C-1. The weight of each can is 60 g and it contains 0.45 kg of pea soup. Assume that the heat content of the autoclave walls above 40°C is 1.6 x 10 4 kJ and that there is no heat loss through the walls, and the specific heat for water is 4.2 kJ kg-1 °C-1.

Question

Engineering

Dominick Harrington

29 September, 16:15

An autoclave contains 1000 cans of pea soup. It is heated to an overall temperature of 100°C. If the cans are to be cooled to 40°C before leaving the autoclave, how much cooling water is required if it enters at 15°C and leaves at 35°C? The specific heats of the pea soup and the can metal are respectively 4.1 kJ kg-1 °C-1 and 0.50 kJ kg-1 °C-1. The weight of each can is 60 g and it contains 0.45 kg of pea soup. Assume that the heat content of the autoclave walls above 40°C is 1.6 x 10 4 kJ and that there is no heat loss through the walls, and the specific heat for water is 4.2 kJ kg-1 °C-1.

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Tater · Answer 1

1529.76 kg of cooling water

Explanation:

Heat gained by cooling water = Total heat loss by autoclave, pepper soup and can

Heat gained by cooling water = mC (T2 - T1) = m*4.2*1000 (35 - 15) = 84000m J

Heat loss by autoclave = 1.6*10^4 kJ = 1.6*10^4 * 1000 = 1.6*10^7 J

Heat loss by pepper soup = mC (T2 - T1) = 0.45*1000*4.1*1000 (100 - 40) = 0.45*1000*4.1*1000*60 = 1.107*10^8 J

Heat loss by can = mC (T2 - T1) = 60/1000 * 1000 * 0.5*1000 (100 - 40) = 60*0.5*1000*60 = 1.8*10^6 J

Total heat loss = (1.6*10^7) + (1.107*10^8) + (1.8*10^6) = 1.285*10^8 J

84000m = 1.285*10^8

m = (1.285*10^8) / 84000 = 1529.76 kg of cooling water