How much work must be done to extract 10.0 J of heat (a) from a reservior at 7ºC and transfer it to one at 27ºC by means of a refrigerator using a Carnot cycle; (b) from one at -73ºC to one at 27ºC; (c) from one at -173ºC to one at 27ºC; and (d) from one at -223ºC to one at 27ºC?
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(a) To obtain work W, substitute 10.0 J for QL, 27° C for TH and 7° C for TL in the equation W = QL (TH/ TL – 1),
W = QL (TH/ TL – 1)
= 10.0 J (27° C/ 7° C -1)
= 10.0 J ((27+273) K /(7+273) K -1)
= 10.0 J (300 K/280 K – 1)
= 0.714 J
Therefore the work done would be 0.714 J.
(b) To obtain work W, substitute 10.0 J for QL, 27° C for TH and -73° C for TL in the equation W = QL (TH/ TL – 1),
W = QL (TH/ TL – 1)
= 10.0 J (27° C/ (-73° C) -1)
= 10.0 J ((27+273) K /(-73+273) K -1)
= 10.0 J (300 K/200 K – 1)
= 5.00 J
Therefore the work done would be 5.00 J.
(c) To obtain work W, substitute 10.0 J for QL, 27° C for TH and -173° C for TL in the equation W = QL (TH/ TL – 1),
W = QL (TH/ TL – 1)
= 10.0 J (27° C/ (-173° C) -1)
= 10.0 J ((27+273) K /(-173+273) K -1)
= 10.0 J (300 K/100 K – 1)
= 20.0 J
Therefore the work done would be 20.0 J.
d) To obtain work W, substitute 10.0 J for QL, 27° C for TH and -223° C for TL in the equation W = QL (TH/ TL – 1),
W = QL (TH/ TL – 1)
= 10.0 J (27° C/ (-223° C) -1)
= 10.0 J ((27+273) K /(-223+273) K -1)
= 10.0 J (300 K/50 K – 1)
= 50.0 J
Therefore the work done would be 50.0 J.
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