A 307-g sample of an unknown mineral was heated to 98.7°C and placed into a calorimeter containing 72.4 g of water at 23.6°C. The heat capacity of the calorimeter was 15.7 J/°C. The final temperature in the calorimeter was 32.4°C. What is the specific heat capacity of the mineral?
A) 0.124 J/g·°C
B) 0.131 J/g·°C
C) 0.138 J/g·°C
D) 0.145 J/g·°C
E) None of these choices is correct.

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat transferred from the mineral to the water and calorimeter:
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. 
We can assume that the heat lost by the mineral is equal to the heat gained by the water and calorimeter:
-q = qwater+qcalorimeter

-q = (mwater x cwater x ΔTwater) + (Ccalorimeter x ΔTcalorimeter)
where Ccalorimeter is the heat capacity of the calorimeter.

Substituting the given values, we get: 
- q = (72.4 g) (4.18 J/g·°C) (32.4°C - 23.6°C) + (15.7 J/°C) (32.4°C - 23.6°C)
q = - (30,591.36 J)

Since the mineral lost heat to the water and calorimeter, its q value is negative.

Next, we can calculate the specific heat capacity of the mineral:

q = mcΔT

- (30,591.36 J) = (0.307 kg) c (98.7°C - 32.4°C)

c = 0.138 J/g·°C.

Therefore, the answer is C.

A 307-G Sample of an Unknown Mineral Was Heated to 98.7°C