Question

What is the entropy change in the surroundings when one mole of ice melts at 0.00°c in a large room maintained at 32.0°c? the heat of fusion (melting) of ice is 6.01 kj/mol.

Answer 1

The entropy change in the surroundings when one mole of ice melts at 0.00°C in a 32.0°C room is -19.7 J/mol·K, indicating the surroundings are losing heat.

Step-by-step explanation:

The question asks about the entropy change of the surroundings when one mole of ice melts at 0.00°C in a room that is at 32.0°C. The enthalpy of fusion of ice is 6.01 kJ/mol. To calculate the entropy change of the surroundings (ΔSsurroundings), we use the heat transferred (Q), which equals the enthalpy of fusion, and the temperature of the surroundings (T) in Kelvin.

First, we convert the room temperature from Celsius to Kelvin: T = 32.0°C + 273.15 = 305.15 K. Then, we calculate the entropy change of the surroundings using the formula:
ΔSsurroundings = -Q/T.

Since the entropy change is related to the surroundings and the process is endothermic, we expect the surroundings to lose heat, and the sign of the change in entropy will be negative as the formula indicates.

ΔSsurroundings = -6.01 kJ/mol / 305.15 K
ΔSsurroundings = -0.0197 kJ/mol·K or -19.7 J/mol·K when we convert kJ to J by multiplying by 1,000.