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 specific heat of the alcohol is closest to
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

In the first run, the heat lost by the alcohol is gained by the water:
$$Q_{lost\ by\ alcohol}=Q_{gained\ by\ water}$$
$$(m_{alcohol}	imes C_{alcohol}	imes \Delta T)=(m_{water}	imes C_{water}	imes \Delta T)$$
$$(0.2\ kg)	imes C_{alcohol}	imes (15\ ^{\circ}C)= (0.3\ kg)	imes (4190\ J/kg	imes K)	imes (15\ ^{\circ}C)$$ 
$$C_{alcohol}=2100\ J/kg	imes K$$

Similarly, in the second run:
$$(m_{alcohol}	imes C_{alcohol}	imes \Delta T)=(m_{water}	imes C_{water}	imes \Delta T)$$
$$(0.2\ kg)	imes C_{alcohol}	imes (10\ ^{\circ}C)= (0.5\ kg)	imes (4190\ J/kg	imes K)	imes (10\ ^{\circ}C)$$ 
$$C_{alcohol}=1900\ J/kg	imes K$$
Taking the average of the two values we get:
$$C_{alcohol}=(1900+2100)/2=2000\ J/kg	imes K$$
Therefore, the closest option is (C) 2100 J/kg • K.

Two Experimental Runs Are Performed to Determine the Calorimetric Properties