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 time interval mentioned in the introduction, find the entropy change of the boiling water?
B)For the time interval mentioned in the introduction, find the entropy change of the ice-water mixture?
C)For the time interval mentioned in the introduction, find the entropy change of the copper rod?
D)For the time interval mentioned in the introduction, find the total entropy change of the entire system?
To find the entropy change of each component, we need to use the equation:
ΔS = Q / T
where ΔS is the entropy change, Q is the heat transfer, and T is the temperature in Kelvin.
A) Entropy change of boiling water:
Since the copper rod is insulated, the only heat transfer is between the boiling water and the ice-water mixture. The heat transfer from the boiling water to the ice-water mixture causes the ice to melt. We can use the heat of fusion equation to find the heat transfer:
Q = m * L
where Q is the heat transfer, m is the mass of the ice melted, and L is the latent heat of fusion for ice (333,000 J/kg).
Given that 0.180 kg of ice melts, the heat transfer is:
Q = 0.180 kg * 333,000 J/kg = 59,940 J
The temperature of the boiling water is 100 degrees Celsius, which is 373 Kelvin. Therefore, the entropy change of the boiling water is:
ΔS = Q / T = 59,940 J / 373 K = 160.73 J/K
B) Entropy change of the ice-water mixture:
The heat transfer during the process of melting the ice is transferred to both the ice and the water. Since the temperatures of both are constant during this process, the entropy change for both components will be the same.
Using the same heat transfer value as in part A:
ΔS = Q / T = 59,940 J / 273 K = 219.51 J/K
C) Entropy change of the copper rod:
In this case, there is no heat transfer, since the rod is insulated. Therefore, the entropy change for the copper rod is zero:
ΔS = 0 J/K
D) Total entropy change of the entire system:
To find the total entropy change, we sum the individual entropy changes of each component:
ΔS_total = ΔS_boiling water + ΔS_ice-water mixture + ΔS_copper rod
ΔS_total = 160.73 J/K + 219.51 J/K + 0 J/K = 380.24 J/K
Therefore, the total entropy change of the entire system is 380.24 J/K.