Question

For the process H2O(s) -> H2O(l), calculate the change in entropy for the universe in kJ/K at 26.2 °C, given that the surroundings transfer 47.7 kJ of heat to the system.

a) -25.4 kJ/K
b) 25.4 kJ/K
c) -47.7 kJ/K
d) 47.7 kJ/K

Answer 1

The change in entropy for the universe in the given process is -47.7 kJ/K.

Step-by-step explanation:

The change in entropy for the universe in the given process can be calculated using the formula: ΔS(universe) = ΔS(system) + ΔS(surroundings). In this case, the surroundings transfer 47.7 kJ of heat to the system. Since the surroundings are transferring heat to the system, the ΔS(surroundings) will be positive. The ΔS(system) can be calculated using the formula: ΔS(system) = q(rev)/T, where q(rev) is the heat transferred in a reversible process and T is the temperature.

Plugging in the values, we get: ΔS(system) = (6.00 kJ)/(26.2 °C + 273.15) = 0.0228 kJ/K.

Since ΔS(system) and ΔS(surroundings) have opposite signs, we have: ΔS(universe) = ΔS(system) + ΔS(surroundings) = 0.0228 kJ/K - 47.7 kJ/K = -47.68 kJ/K.

Therefore, the change in entropy for the universe in this process is approximately -47.7 kJ/K (Option c).