anonymous
  • anonymous
What is the change in entropy for a system that has 250 J of heat added to it at 25°C? -10 J/K, -0.84 J/K, 0.84 J/K, or 10 J/K **thank you!!
Physics
chestercat
  • chestercat
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Michele_Laino
  • Michele_Laino
it is simple, we have to apply this formula: \[\Large \Delta S = \frac{Q}{T}\] where \delta S is the entropy change
anonymous
  • anonymous
okay!! so 250/25 = 10 ? so our solution is 10 J/K ?
Michele_Laino
  • Michele_Laino
no, since T is the absolute temperature

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anonymous
  • anonymous
ohhh i am unsure then :/
Michele_Laino
  • Michele_Laino
we have: \[T = t + 273\] t is the Celsius temperature
anonymous
  • anonymous
ohh so we would get 273+25=298?
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
250/298=0.84? so our solution is 0.84 J/K?
Michele_Laino
  • Michele_Laino
correct!
anonymous
  • anonymous
yay! thank you!! :D
Michele_Laino
  • Michele_Laino
thank you!! :D

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