
#1
Nov912, 01:31 AM

P: 6

If 12007 kJ of heat is lost to the surroundings with an ambient temperature of 25 degrees centigrade during a cooling process, and the ambient temperature of the surroundings is unaffected by the heat addition, what is the entropy change of the surroundings?
If Δs=∫δQ/T, then Δs=ΔQ/T=12007 kJ/(25+273.15)K= 40.272 kJ/K. Is my thinking process here correct? Thanks! 



#2
Nov912, 07:44 PM

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