Low temperature research is often performed on samples whose temperatures are on
ID: 3162808 • Letter: L
Question
Low temperature research is often performed on samples whose temperatures are on the order 0.001 degree K = 1 millidegree K. (a) The price of work (in the form, say, of electrical energy) is roughly 25 cents per kilowatt-hour. What would be the minimum cost of extracting 1 calorie (=4.18 J) of heat from a system at 1 millidegree K if the surrounding atmosphere is at 290 degree K? (b) The cost of extracting one calorie in part (a) may not seem very high, but imagine that it cost you that much for each calorie you extract from inside your refrigerator. Estimate, very roughly, what your monthly electrical bill would be in that case. You might start by estimating the total amount of liquids (soft drinks, juice, milk, beer, ...) you cool down from room temperature to ~ 40 degree F in a month, and how much ice you freeze in a month. This should convince you that low temperature experiments must be done on small, well insulated samples.Explanation / Answer
It is working of a heat pump reverse of a heat engine.
It takes a nan amount heat q2 from the cold body at temperature t1 and rejects it to a hot body at temprature T2
The work done
W = Q2 (T1/T2 -1)
Q2 = 4.18 J, T1 = 0.001K, T2 = 273 +290 = 563 K
W = 4.18(0.001/563 -1) = -4.18 J
-ve sign indicates work is done on the system to extract heat from a cold body
cost = 4.18 *25/1000*3600 = 2.90e-5 cents
b) for refrigerator we assume a volume of 500 Ltrs average
half of it filled = 250 Ltrs
we assume the eat capcity of the material inside equal to that of water
waterre heat capacity = 4186 J/kg-K
approximately 250 kg of material inside the refrigirator
to be colled from 30 C to 4 C
heat to be removed = 250*4186*26
= 27209 kJ
= 7.56 kwh
cost = 7.56*25 = 1.89$
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