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
To calculate the entropy change to the surroundings when 1 mol of ice melts, we need to use the equation:
ΔSsurroundings = -ΔH / T
Where:
- ΔSsurroundings is the entropy change to the surroundings
- ΔH is the heat of fusion of ice
- T is the temperature in Kelvin
First, we need to convert the hand temperature from Celsius to Kelvin:
T = 32°C + 273.15 = 305.15 K
Next, we can substitute the values into the equation:
ΔH = 6.01 kJ/mol = 6010 J/mol
T = 305.15 K
ΔSsurroundings = - (6010 J/mol) / 305.15 K
Calculating the value:
ΔSsurroundings = -19.7 J/K
The entropy change to the surroundings when 1 mol of ice melts in someone's hand is approximately -19.7 J/K.
Therefore, the correct answer is c. -19.7 J/K.
To find the entropy change to the surroundings, we can use the equation:
ΔS_system + ΔS_surroundings = ΔS_total
ΔS_system represents the entropy change of the system, which in this case is the melting of 1 mol of ice. Since ice is transitioning from a solid to a liquid state, the entropy change can be found using the equation:
ΔS_system = ΔH_fusion / T
Where ΔH_fusion is the heat of fusion of ice (6.01 kJ/mol) and T is the final temperature of the water (0°C or 273 K).
Plugging in the values, we get:
ΔS_system = 6.01 kJ/mol / 273 K
Next, we can find the entropy change of the surroundings, which can be calculated using the equation:
ΔS_surroundings = - ΔH_fusion / T
Since the heat is transferred from the surroundings to the system, the entropy change of the surroundings is the negative of the entropy change of the system.
Plugging in the values, we get:
ΔS_surroundings = - 6.01 kJ/mol / 273 K
Converting the units from kJ to J, we have:
ΔS_surroundings = - 6010 J/mol / 273 K
To find the entropy change to the surroundings when 1 mol of ice melts in someone's hand, we need to divide this value by the number of moles of ice:
ΔS_surroundings = - 6010 J/mol / 273 K * (1 mol / 1 mol)
Simplifying this expression, we get:
ΔS_surroundings = - 6010 J/K
The correct answer is b) -22.0 J/K.