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. +19.7 J/K
b. –22.0 J/K e. +188 J/K
c. –19.7 J/K
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. +19.7 J/K
b. –22.0 J/K e. +188 J/K
c. –19.7 J/K
dSsurr = -dH/t -6010/305 = ? J/K
To calculate the entropy change to the surroundings when 1 mol of ice melts in someone's hand, we can use the formula:
ΔS_surroundings = -q_surroundings / T
where ΔS_surroundings is the entropy change of the surroundings, q_surroundings is the heat transfer to the surroundings, and T is the temperature in Kelvin.
First, let's calculate the heat transfer to the surroundings using the given heat of fusion of ice:
q_surroundings = ΔH_fusion * n
q_surroundings = 6.01 kJ/mol * 1 mol
q_surroundings = 6.01 kJ
Next, we need to convert the final temperature of the surroundings from Celsius to Kelvin:
T = 32°C + 273.15
T = 305.15 K
Now, we can substitute the values into the formula to calculate the entropy change of the surroundings:
ΔS_surroundings = -(6.01 kJ) / (305.15 K)
ΔS_surroundings ≈ -0.0197 kJ/K ≈ -19.7 J/K
Therefore, the entropy change to the surroundings when 1 mol of ice melts in someone's hand is approximately -19.7 J/K. The correct answer is c. –19.7 J/K.
To find the entropy change to the surroundings when 1 mol of ice melts in someone's hand, we can use the equation:
ΔS = q/T
where ΔS is the entropy change, q is the heat absorbed or released, and T is the temperature in Kelvin.
First, let's calculate the heat absorbed during the melting of 1 mol of ice. The heat of fusion of ice is given as 6.01 kJ/mol. However, we need to convert this value to joules:
Heat of fusion = 6.01 kJ/mol = 6.01 * 10^3 J/mol
Next, we need to calculate the temperature change of the surroundings. The initial temperature of the hand is given as 32°C. We need to convert this to Kelvin by adding 273.15:
Initial temperature = 32°C + 273.15 = 305.15 K
The final temperature of the water is given as 0°C, which is equal to 273.15 K.
Now, we can substitute the values into the equation:
ΔS = q/T = (6.01 * 10^3 J/mol) / (305.15 K - 273.15 K)
Simplifying the equation:
ΔS = (6.01 * 10^3 J/mol) / 32 K
Calculating the value:
ΔS = 188 J/K
Therefore, the entropy change to the surroundings when 1 mol of ice melts in someone's hand is +188 J/K (option e).