Suppose that you want to freeze 1 kg of water for a party, and only have 5 minut

Question

Suppose that you want to freeze 1 kg of water for a party, and only have 5 minutes to do it. Relying on nothing but your keen physical abilities, a dixicup, 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 350 K. Since you are a brilliant UIUC student, assume that your refrigerator has the maximum possible efficiency. The Latent Heat of Fusion of water is 3.33×10?5 J/kg
Assume that the initial temperature of the water is 273 K.

1. What is the change (including the sign) in entropy in the water per second in J/(Ks)?
2. What is the heat flow ( Q_hot/time ) into the Hot well per second in J/s?
3. How much POWER must you provide to drive your refrigerator in Watt W? (That is work done per second?)

Explanation / Answer

1)
Qc = m*Hfus
Qc = 1*333
Qc = 333 kJ

Qc_dot = Qc/t = 333/300
Qc_dot = 1.11 kW

?S = -Qc_dot/Tfus
?S = -1110/273
?S = -4.065934 W/K = -4.065934 J/Ks

2)
? = Tc/Th-Tc = 273/350 - 273
? = 3.54545

? = Qc/Qh - Qc
3.54545 = 1.11 / Qh - 1.11
3.54545Qh - 3.9354495 = 1.11
Qh = 1.4231 kW = 1423.1 J/s

3)
W = Qh - Qc
W = 1423.1 - 1110
W = 313.1 W

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