What is the entropy change in the surroundings when one mole of ice melts at 0.00 $\circ$ C in a large room maintained at 32.0 $\circ$ C? The heat of fusion of ice is 6.01 kJ/mol.

Question

Quizplus · Accepted Answer

We can use the formula for entropy change in surroundings:

ΔS surr = -ΔH/T

where ΔH is the enthalpy change (heat) absorbed or released by the system, and T is the temperature of the surroundings.

In this case, one mole of ice is melting, so the heat absorbed by the system (ice) is:

ΔH = 6.01 kJ/mol

The temperature of the surroundings is:

T = 32.0°C = 305.15 K

Note that we need to convert the temperature from Celsius to Kelvin to use it in the formula.

Now we can calculate the entropy change in surroundings:

ΔS surr = -ΔH/T
ΔS surr = -(6010 J/mol) / (305.15 K)
ΔS surr = -19.7 J/K

The negative sign indicates that the surroundings are losing entropy (i.e. becoming more ordered), which is expected since heat is being absorbed by the system.

Therefore, the answer is C) -19.7 J/K.

What Is the Entropy Change in the Surroundings When One ∘\circ∘

What Is the Entropy Change in the Surroundings When One $\circ$