Question

Two hundred grams of water at 0°C is brought into contact with a heat reservoir at 80°C. After thermal equilibrium is reached, what is the temperature of the water? Of the reservoir? How much heat has been transferred in the process? What is the entropy change of the water? Of the reservoir? What is the entropy change of the universe?

a) 40°C, 80°C, 8.3 kJ, 83.2 J/K, -83.2 J/K, -
b) 40°C, 80°C, 8.3 kJ, -83.2 J/K, 83.2 J/K, 83.2 J/K
c) 60°C, 80°C, 8.3 kJ, 83.2 J/K, -83.2 J/K, 83.2 J/K
d) 60°C, 80°C, 8.3 kJ, -83.2 J/K, 83.2 J/K, -

Answer 1

After reaching thermal equilibrium with a heat reservoir at 80°C, the temperature of the water is 40°C. The temperature of the reservoir remains constant at 80°C. The amount of heat transferred in the process is 8.36 kJ, and the entropy change of the water, reservoir, and the universe are all 26.68 J/K.

Step-by-step explanation:

The temperature of the water after reaching thermal equilibrium with the heat reservoir at 80°C can be determined using the principle of heat transfer. The heat gained by the water is equal to the heat lost by the reservoir. Assuming no heat is lost to the surroundings, we can use the equation:

mcΔT = mcΔT
(200g)(4.18 J/g°C)(T - 0°C) = (200g)(4.18 J/g°C)(80°C - T)

Solving this equation, we find that the temperature of the water at equilibrium is 40°C. The temperature of the reservoir remains constant at 80°C as it is an idealized heat reservoir. Heat transferred in the process can be calculated using the equation: Q = mcΔT.

Thus, the amount of heat transferred is (200g)(4.18 J/g°C)(40°C) = 8.36 kJ. The entropy change of the water can be calculated using the equation: ΔS = Q/T. Therefore, the entropy change of the water is (8.36 kJ)/(313K) = 26.68 J/K. The entropy change of the reservoir can be calculated in the same way and is also found to be 26.68 J/K. Finally, since the system is closed, the entropy change of the universe is the sum of the entropy changes of the water and the reservoir, resulting in a value of 53.36 J/K.