Entropy: Heat addition to surrounding.

AI Thread Summary
The discussion revolves around calculating the entropy change of the surroundings when 12007 kJ of heat is lost during a cooling process at an ambient temperature of 25 degrees Celsius. The formula used is Δs=ΔQ/T, where ΔQ is the heat lost and T is the absolute temperature in Kelvin. The calculation yields an entropy change of approximately 40.272 kJ/K. The reasoning and calculation presented are confirmed to be correct. This highlights the relationship between heat loss and entropy change in thermodynamic processes.
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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!
 
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afpskierx said:
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!
Yes.
 

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