Entropy: Heat addition to surrounding.

  1. 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. jcsd
  3. Chestermiller

    Staff: Mentor

    Yes.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook