What is the entropy change to the surroundings when 1 mol of ice melts in someone’s hand if the hand temperature is 32°C? Assume a final temperature for the water of 0°C. The heat of fusion of ice is 6.01 kJ/mol.

a. –188 J/K d.
b. –22.0 J/K
c. –19.7 J/K
d. +19.7 J/K
e. +188 J/K

dSsurr= -dH/T = -(6010/305.15) = ?J/K

To find the entropy change to the surroundings when 1 mol of ice melts in someone's hand, we need to calculate the heat transferred from the surroundings to the ice.

The heat transferred (q) can be calculated using the equation:
q = n * ΔHf

Where:
q = heat transferred
n = number of moles of ice
ΔHf = heat of fusion of ice

Given:
n = 1 mol
ΔHf = 6.01 kJ/mol

To convert kJ to J, we multiply by 1000:
ΔHf = 6.01 * 1000 J/mol = 6010 J/mol

Now we need to calculate the change in entropy (ΔS) using the equation:
ΔS = q / T

Where:
ΔS = change in entropy
q = heat transferred
T = temperature

Given:
T = 32°C = 32 + 273.15 = 305.15 K

Plugging in the values, we get:
ΔS = 6010 J/mol / 305.15 K = 19.68 J/K

Since the question asks for the entropy change to the surroundings, the answer is +19.7 J/K (option d).

To determine the entropy change to the surroundings when 1 mol of ice melts in someone's hand, we need to calculate the change in entropy formula:

ΔS = q/T

Where:
ΔS = Change in entropy
q = Heat transferred
T = Temperature

First, let's calculate the heat transferred during the phase change. The heat of fusion of ice is given as 6.01 kJ/mol. Since 1 kJ is equal to 1000 J, the heat of fusion is 6.01 × 1000 J/mol.

Next, we need to calculate the temperature difference. The starting temperature is given as 32°C, and the final temperature is 0°C. To ensure the temperatures are in Kelvin, we need to add 273.15 to each value, converting them to Kelvin. Therefore, the starting temperature is 305.15 K and the final temperature is 273.15 K.

Now we can substitute the values into the formula:

ΔS = (6.01 × 1000 J/mol) / (273.15 K)

Calculating this gives us:

ΔS ≈ 22 J/K

Since the question asks for the entropy change to the surroundings, the answer is the negative of this value:

ΔS ≈ -22 J/K

Therefore, the correct choice among the given options is b. –22.0 J/K.