Two experimental runs are performed to determine the calorimetric properties of an alcohol that has a melting point of -10°C. In the first run, a 200-g cube of frozen alcohol, at the melting point, is added to 300 g of water at 20°C in a styrofoam container. When thermal equilibrium is reached, the alcohol-water solution is at a temperature of 5.0°C. In the second run, an identical cube of alcohol is added to 500 g of water at 20°C and the temperature at thermal equilibrium is 10°C. The specific heat of water is 4190 J/kg·K. Assume that no heat is exchanged with the styrofoam container and with the surroundings. The heat of fusion of the alcohol is closest to A) 5.5 × 10⁴ J/kg B) 6.3 × 10⁴ J/kg C) 7.1 × 10⁴ J/kg D) 7.9 × 10⁴J/kg E) 8.7 × 10⁴ J/kg

Question

Quizplus · Accepted Answer

First, we can find the heat released by the alcohol to cool down to the final temperature using the specific heat of water:
Q = mcΔT
For the first run: 
Q1 = (0.2 kg)(2.51 kJ/kg·K)(-10°C - 5.0°C) = -6.3 kJ
For the second run: 
Q2 = (0.2 kg)(2.51 kJ/kg·K)(-10°C - 10°C) = -10.04 kJ
Next, we can use the heat released in each run to find the heat of fusion (ΔHf) of the alcohol using the equation:
ΔHf = -Q/m
where m is the mass of the alcohol used (0.2 kg).
For the first run: 
ΔHf1 = -(-6.3 kJ) / 0.2 kg = 31.5 kJ/kg
For the second run: 
ΔHf2 = -(-10.04 kJ) / 0.2 kg = 50.2 kJ/kg
The average of the two values is closest to the given answer B:
(31.5 kJ/kg + 50.2 kJ/kg) / 2 ≈ 40.85 kJ/kg ≈ 6.3 × 10^4 J/kg

Two Experimental Runs Are Performed to Determine the Calorimetric Properties