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 × 10⁴ J/kg. E) 8.7 × 10⁴ J/kg.

Question

Quizplus · Accepted Answer

First, we can calculate the heat transferred from the alcohol to water in each experiment using: 
Q = mcΔT 
where Q is the heat transferred, m is the mass of water, c is the specific heat of water, and ΔT is the change in temperature of the water.

For the first experiment: 
Q = (300 g)(4190 J/kg K)(5.0°C - 20°C) 
Q = -2,507,400 J

For the second experiment: 
Q = (500 g)(4190 J/kg K)(10°C - 20°C) 
Q = -2,095,000 J

In both experiments, the heat transferred from the alcohol to the water was equal to the heat required to melt the 200 g cube of alcohol:

Q = mL 
where m is the mass of the alcohol and L is the heat of fusion of the alcohol.

We can set up two equations based on this:

-2,507,400 J = (200 g)L 
-2,095,000 J = (200 g)L

Solving for L in each equation gives:

L = 6.3 × 10^4 J/kg

Therefore, the heat of fusion of the alcohol is closest to option B, 6.3 × 10^4 J/kg.

Two Experimental Runs Are Performed to Determine the Calorimetric Properties