Thermodynamics entropy problem

  • #1
200
0

Homework Statement


Find (a) the energy absorbed as heat and (b) the change in entropy of a 2.00 kg block of copper whose
temperature is increased reversibly from 25°C to 100°C. The specific heat of copper is 386 J/kg K .


Homework Equations



dS=dQ/T
dS=nRln(final volume/initial volume) **for an ideal gas**

The Attempt at a Solution

\

Well I got the first part. The energy is just
dQ=cmdT
which is (386)(2)(75K)= 57900J

however, when i got to the second part, I didn't know what to do since the copper isn't an ideal gas, and it has a non-constant temperature so I can't use the first equation. Any help would be appreciated!!!
 

Answers and Replies

  • #2
200
0
Entropy Thermodynamics problem

Homework Statement



Find (a) the energy absorbed as heat and (b) the change in entropy of a 2.00 kg block of copper whose
temperature is increased reversibly from 25°C to 100°C. The specific heat of copper is 386 J/kg K .

Homework Equations



dS=dQ/T
dS=nr ln(vf/vi) **for an ideal gas**

The Attempt at a Solution



The first part was relatively easy. I just used the equation:
dQ=cmdT
so, dQ=(386)(2)(75)=57900 J

But for the 2nd part, I dont know what to do because copper is not an ideal gas, ruling out the 2nd equation, and the temperature is not constant, meaning I cannot use the first equation. Any help would be great!!!
 
  • #3
1,187
5


dS = dQ/T
dQ = cmdT
dS = cmdT/T
Integrate both sides to get delta(S) = cm*ln(Tf/Ti); be sure to use Kelvin.
Its been a while since I've done thermodynamics, so I might be wrong...though I only applied a bit of mathematical procedure, so it seems as though it would be fine.
 
  • #4
200
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sounds right to me.... thanks for the help!
 

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