Question

307 g of an unknown mineral is heated to 98.7 °C and placed into a calorimeter that contains 72.4 g of water at 23.6 °C. The heat capacity of the empty calorimeter was 15.7 J/K. The final temperature of the calorimeter was 32.4 °C. What is the specific heat capacity of the mineral in J/gK?

Answer 1

Answer:

0.138 J/g.K

Explanation:

According to the law of conservation of energy, the sum of the heat released bt he mineral and the heats absorbed by the water and the calorimeter is equal to zero.

Qm + Qw + Qc = 0

Qm = - Qw - Qc   [1]

The heat absorbed by the calorimeter can be calculated using the following expression.

Q = C . ΔT

where,

C is the heat capacity

ΔT is the change in temperature

The heat released by the mineral and the one absorbed by the water can be calculated using the following expression.

Q = c . m . ΔT

where,

c is the specific heat capacity

m is the mass

ΔT is the change in temperature

Replacing in [1],

$Qm = -Qw - Qc\\c_{m}.m_{m}.\Delta T_{m}=-c_{w}.m_{w}.\Delta T_{w}-C_{c}.\Delta T_{c}\\c_{m}=\frac{-c_{w}.m_{w}.\Delta T_{w}-C_{c}.\Delta T_{c}}{m_{m}.\Delta T_{m}} \\c_{m}=\frac{-(4.18J/g.K).(72.4g).(305.4K-296.6K)-(15.7J/K).(305.4K-296.6K) }{(307g).(305.4K-371.7K)} \\c_{m}=0.138J/g.K$