One end of a copper rod is immersed in boiling water at 100 degrees celsius and the other end in an ice-water mixture at 0 degrees celsius. The sides of the rod are insulated. After steady-state conditions have been achieved in the rod, 0.180 kg of ice melts in a certain time interval.
A)For the given time interval, find the entropy change of the boiling water?
B)Find the entropy change of the ice-water mixture?
C)Find the entropy change of the copper rod?
D)Find the total entropy change of the entire system?
To find the entropy change of each component of the system, we need to calculate the heat transferred and the temperature change for each component.
A) Entropy change of the boiling water:
In this case, the boiling water is at a constant temperature of 100 degrees Celsius. Since the temperature is constant, there is no temperature change. The only factor contributing to the entropy change is the heat transferred.
To find the heat transferred, we can use the equation:
Q = m * c * ∆T
Where:
Q is the heat transferred,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
∆T is the temperature change.
In this case, the temperature change is zero, so the equation simplifies to:
Q = m * c * 0 = 0
Since there is no heat transferred, the entropy change of the boiling water is zero.
B) Entropy change of the ice-water mixture:
In this case, the ice-water mixture is at a constant temperature of 0 degrees Celsius. Similarly to the boiling water, since the temperature is constant, there is no temperature change. Therefore, there is no change in entropy for the ice-water mixture.
C) Entropy change of the copper rod:
To calculate the entropy change of the copper rod, we need to consider the heat transferred and the temperature change.
First, let's find the heat transferred from the boiling water to the copper rod. We can use the same equation as before:
Q = m * c * ∆T
In this case, we know the mass of the ice melted, which is 0.180 kg. Moreover, the specific heat capacity of water is approximately 4186 J/kg°C, and the temperature change is 100°C. Substituting the values into the equation, we can find the heat transferred.
Q = (0.180 kg) * (4186 J/kg°C) * (100°C) = 75012 J
The same amount of heat is transferred from the copper rod to the ice-water mixture, as heat transfer is symmetric. Therefore, the heat transferred from the copper rod is also 75012 J.
To find the entropy change of the copper rod, we can use the formula:
∆S = Q / T
Where:
∆S is the entropy change,
Q is the heat transferred, and
T is the temperature of the component.
Since the temperature change of the copper rod is 100°C, we convert it to Kelvin by adding 273.15.
T = 100°C + 273.15 = 373.15 K
∆S = (75012 J) / (373.15 K) ≈ 201.13 J/K
Therefore, the entropy change of the copper rod is approximately 201.13 J/K.
D) Total entropy change of the entire system:
The total entropy change of the entire system is the sum of the entropy changes of the individual components. In this case, since the entropy changes of the boiling water and ice-water mixture are zero, the total entropy change is equal to the entropy change of the copper rod:
Total entropy change = Entropy change of the copper rod ≈ 201.13 J/K.