Two experimental runs are performed to determine the calorimetric properties of an alcohol which 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°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 capacity of water is 4190 J/kg ∙ K.Assume no heat is exchanged with the styrofoam container and the surroundings.What is the specific heat capacity of the alcohol?
A)1700 J/kg ∙ K
B)1900 J/kg ∙ K
C)2100 J/kg ∙ K
D)2300 J/kg ∙ K
E)2500 J/kg ∙ K

Question

Quizplus · Accepted Answer

We can use the formula for heat transfer in calorimetry:

Q = mcΔT

where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

In the first run, the heat transferred from the alcohol to the water is:

Q1 = mcΔT1 = (200 g) c (5°C - (-10°C)) = 3000c J

In the second run, the heat transferred from the alcohol to the water is:

Q2 = mcΔT2 = (200 g) c (10°C - (-10°C)) = 6000c J

The heat transferred must be the same in both runs, so we can set Q1 = Q2:

3000c = 6000c/2

c = 2100 J/kg ∙ K

Therefore, the specific heat capacity of the alcohol is 2100 J/kg ∙ K, which is choice C.

Two Experimental Runs Are Performed to Determine the Calorimetric Properties