Question

A 100.0 kg block of aluminum (Body A), is initially @ 100°C. A 209.3 kg block of steel (Body B) is initially at 0°C. The two blocks are then connected via a short, copper cylinder for 1.5 minutes, during which time heat flows at a rate of 10.0 W. Other than when the connection occurs, the blocks are perfectly insulated.\ $C_{AI} = 0.90$ kJ/kg-K $C_{steel} = 0.43$ kJ/kg-K a) Determine the amount of entropy produced in this process. b) Redo Part (a) treating each body as a thermal energy reservoir. Is the answer close to that of Part (a)? Explain why or why not. Hint: For the thermal energy reservoirs (TERs), use the definition of entropy change for an internally reversible process (i.e., since the TERs are each always at a uniform temperature, they, by definition, remain in thermal equilibrium). Alternatively, you can use the equation: $S_2 - S_1 = mcln(\frac{T_2}{T_1})$ However, you have to solve the temperature rise for a finite mass as a parameter and then take the limit of $mcln(\frac{T_2}{T_1})$ as the mass goes to infinity (and the corresponding temperature rise goes to zero). It works out nicely!

          A 100.0 kg block of aluminum (Body A), is initially @ 100°C. A 209.3 kg block of steel (Body B) is initially at 0°C. The two blocks are then connected via a short, copper cylinder for 1.5 minutes, during which time heat flows at a rate of 10.0 W. Other than when the connection occurs, the blocks are perfectly insulated.\
$C_{AI} = 0.90$ kJ/kg-K
$C_{steel} = 0.43$ kJ/kg-K
a) Determine the amount of entropy produced in this process.
b) Redo Part (a) treating each body as a thermal energy reservoir. Is the answer close to that of Part (a)? Explain why or why not.
Hint: For the thermal energy reservoirs (TERs), use the definition of entropy change for an internally reversible process (i.e., since the TERs are each always at a uniform temperature, they, by definition, remain in thermal equilibrium).
Alternatively, you can use the equation:
$S_2 - S_1 = mcln(\frac{T_2}{T_1})$
However, you have to solve the temperature rise for a finite mass as a parameter and then take the limit of $mcln(\frac{T_2}{T_1})$ as the mass goes to infinity (and the corresponding temperature rise goes to zero). It works out nicely!

A 100.0 kg block of aluminum (Body A), is initially @ 100°C. A 209.3 kg block of steel (Body B) is initially at 0°C. The two blocks are then connected via a short, copper cylinder for 1.5 minutes, during which time heat flows at a rate of 10.0 W. Other than when the connection occurs, the blocks are perfectly insulated.CAI = 0.90 kJ/kg-K
Csteel = 0.43 kJ/kg-K
a) Determine the amount of entropy produced in this process.
b) Redo Part (a) treating each body as a thermal energy reservoir. Is the answer close to that of Part (a)? Explain why or why not.
Hint: For the thermal energy reservoirs (TERs), use the definition of entropy change for an internally reversible process (i.e., since the TERs are each always at a uniform temperature, they, by definition, remain in thermal equilibrium).
Alternatively, you can use the equation:
S2 - S1 = mcln((T2)/(T1))
However, you have to solve the temperature rise for a finite mass as a parameter and then take the limit of mcln((T2)/(T1)) as the mass goes to infinity (and the corresponding temperature rise goes to zero). It works out nicely!

Added by Bryan V.

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