Entropy: Heat addition to surrounding.
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?
Re: Entropy: Heat addition to surrounding.
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