Physics
posted by BG on .
Refrigerator
Suppose that you want to freeze 1.2 kg of water for a party, and only have 10 minutes to do it. Relying on nothing but your keen physical abilities, a dixie cup, duct tape and a piece of string, you manage to build a refrigerator unit which can do the job.
The temperature inside the refrigeration unit is 273 K, and the temperature outside is Thot = 301 K. Since you are a brilliant UW student, assume that your refrigerator has the maximum possible efficiency. Assume also that the initial temperature of the water is 273 K. The Latent Heat of Fusion of water is 3.33×105 J/kg.
(a) What is the heat flow ( Qcold / time ) out of the cold well per second in J/s?
Hcold = Qcold / time = J/s *
666 OK
(b) What is the change in entropy in the water per second in J/K s?
ΔS / time = J/K s
431.58 NO
HELP: Change in entropy = Change in energy / Absolute Temperature.
HELP: To get the change in entropy of the water, use the total heat of fusion for the 1.2 kg of water. The water freezes at 273 K, and it all happens over a period of 10×60 seconds.
(c) What is the heat flow ( Qhot / time ) into the hot well per second in J/s?
Hhot = Qhot / time = J/s
HELP: Entropy is a conserved quantity. The change in entropy of the cold well MUST (for a perfect refrigerator) be equal to the negative of the change in entropy of the hot well.
HELP: Entropy = Energy transfer / Temperature. You know how much entropy was transferred, and the temperatures of the hot and cold reservoirs, so you can calculate how much energy was transferred for each reservoir.
(d) How much POWER must you provide to drive your refrigerator, in Watts? (That is, work done per second?)
Power = W
I need help with B onward. I can't seem to figure out the change in entropy

LOL