If 1 mol of ice melts at its melting point of 273 K, the entropy change for the ice is 22.0 J/K. If the ice melts in someone’s hand at 34°C, what is the change in the entropy of the universe? Assume a final temperature for the water of 0°C. The enthalpy of fusion for ice is 6.01 kJ/mol.
a. +19.6 J/K d. –2.4 J/K
b. –19.6 J/K e. +41.5 J/K
c. +2.4 J/K
dSsurr = -dH/T = -6010/307
b. –19.6 J/K e. +41.5 J/K
To find the change in the entropy of the universe, we need to consider both the entropy change of the ice and the entropy change of the surroundings (the person's hand).
The entropy change of the ice can be calculated using the equation:
ΔS = q/T
where ΔS is the entropy change, q is the heat transferred, and T is the temperature. We are given that the entropy change for 1 mol of ice is 22.0 J/K, so the entropy change for the given amount of ice can be found using the molar mass of ice.
The molar mass of ice is approximately 18.02 g/mol, so the entropy change for the ice is:
ΔS_ice = (22.0 J/K) * (18.02 g/mol) = 396.44 J/K
Next, let's calculate the heat transferred (q) during the melting process. The heat transferred can be calculated using the equation:
q = ΔH_fusion * n
where ΔH_fusion is the enthalpy of fusion and n is the number of moles. We are given that the enthalpy of fusion for ice is 6.01 kJ/mol, so the heat transferred can be calculated as follows:
q = (6.01 kJ/mol) * (1 mol) = 6.01 kJ
Let's convert this heat value to joules:
1 kJ = 1000 J, so 6.01 kJ = 6010 J
Now, let's calculate the entropy change of the surroundings (the person's hand). The entropy change of the surroundings can be calculated using the equation:
ΔS_surroundings = -q/T
We are given that the initial temperature is 34°C, which is 307 K. We are also given that the final temperature is 0°C, which is 273 K. Therefore, the temperature change is:
ΔT = T_final - T_initial = 273 K - 307 K = -34 K
Now, let's substitute the values into the equation to calculate the entropy change of the surroundings:
ΔS_surroundings = -(6010 J) / (307 K) = -19.57 J/K
Finally, to find the change in the entropy of the universe, we can add the entropy change of the ice to the entropy change of the surroundings:
ΔS_universe = ΔS_ice + ΔS_surroundings = 396.44 J/K + (-19.57 J/K) = 376.87 J/K
The change in the entropy of the universe is approximately 376.87 J/K.
The closest option available is d. –2.4 J/K.
To find the change in the entropy of the universe, we need to calculate the change in entropy of the surroundings (the person's hand) and the change in entropy of the ice.
The change in entropy of the surroundings can be calculated using the equation:
ΔS_surroundings = -ΔH / T
where ΔH is the enthalpy of fusion for ice and T is the temperature in Kelvin.
Given that the enthalpy of fusion for ice is 6.01 kJ/mol and the temperature is 34°C, we need to convert the temperature to Kelvin:
T = 34 + 273 = 307 K
Substituting the values into the equation:
ΔS_surroundings = -(6.01 kJ/mol) / 307 K
Next, we need to calculate the change in entropy of the ice by using the equation:
ΔS_ice = ΔH / T
Given that the enthalpy of fusion for ice is 6.01 kJ/mol and the temperature is 273 K (melting point of ice), we can calculate:
ΔS_ice = (6.01 kJ/mol) / 273 K
Now, we can calculate the change in the entropy of the universe by summing up the entropy changes of the surroundings and the ice:
ΔS_universe = ΔS_surroundings + ΔS_ice
Substituting the calculated values:
ΔS_universe = -(6.01 kJ/mol) / 307 K + (6.01 kJ/mol) / 273 K
To convert from kJ to J, we multiply by 1000:
ΔS_universe = -(6.01 × 10^3 J/mol) / 307 K + (6.01 × 10^3 J/mol) / 273 K
Simplifying the equation gives us the value of the change in the entropy of the universe.