One end of a metal rod is in contact with a thermal reservoir at 745 K, and the other end is in contact with a thermal reservoir at 105 K. The rod and reservoirs make up a closed system. If 7030 J are conducted from one end of the rod to the other uniformly (no change in temperature along the rod) ...
a) ... what is the change in entropy of the 745-K reservoir?
b) ... what is the change in entropy of the 105-K reservoir?
c) ... what is the change in entropy of the rod?
d) ... what is the change in entropy of the system?
To calculate the change in entropy for each component, we can use the equation:
ΔS = Q/T
where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature at which the heat transfer occurs.
a) To find the change in entropy of the 745 K reservoir, we need to know the heat transferred to or from the reservoir. In this case, we know that 7030 J are conducted from one end of the rod to the other uniformly. Since heat is leaving the reservoir, the value of Q would be -7030 J (negative sign indicates heat leaving).
Using the given temperature T = 745 K and the formula, we can calculate:
ΔS = Q/T = (-7030 J) / (745 K) ≈ -9.42 J/K
So, the change in entropy of the 745 K reservoir is approximately -9.42 J/K.
b) Similarly, to find the change in entropy of the 105 K reservoir, we need to know the heat transferred to or from the reservoir. Since 7030 J are conducted from one end of the rod to the other, the heat received by the 105 K reservoir will be 7030 J (positive sign indicates heat received).
Using the given temperature T = 105 K and the formula, we can calculate:
ΔS = Q/T = (7030 J) / (105 K) ≈ 66.95 J/K
So, the change in entropy of the 105 K reservoir is approximately 66.95 J/K.
c) To find the change in entropy of the metal rod, we need to consider the heat transferred within it. Since the temperature is not changing along the rod (uniform temperature gradient), the heat transfer is the same at every point.
Using the same formula ΔS = Q/T, we can calculate:
ΔS = (7030 J + (-7030 J)) / (745 K) ≈ 0
Note that since the heat transferred into one end of the rod equals the heat transferred out of the other end, there is no net change in entropy for the rod.
d) To find the change in entropy of the system, we need to sum up the changes in entropy of all the components. In this case, we have two reservoirs and a rod.
ΔS system = ΔS 745K reservoir + ΔS 105K reservoir + ΔS rod
Since we already calculated the values for the reservoirs and found that ΔS rod is approximately 0, we can plug in the values:
ΔS system = -9.42 J/K + 66.95 J/K + 0
ΔS system ≈ 57.53 J/K
So, the change in entropy of the system is approximately 57.53 J/K.