Question

Fuel Gasoline Song Mass 1g 2g 38 Burn Calculate the heat capacity of the bomb calorimeter in kilojoules per degree Celsius (kJ/°C). The fuel value of each fu provided in the table. Fuel wood coal Fuel value (kJ/g) 18.9 34.6 46.5 29.7 gasoline ethanol

at 1 gram

Wood temp change- 1.9

Coal temp change- 3.5

Gasoline temp change- 4.7

Ethanol temp change- 3.0

Answer 1

In bomb calorimeter, the heat released after burning increases the temperature of the calorimeter.

The heat released (Q) is given by,

$Q=H_V\times m_f$

Here, m_f is the mass of fuel burned and H_v is the heat value of fuel.

Further, this heat raises the temperature of the bomb calorimeter according to the formula,

$Q=C\Delta T\\ \Rightarrow C=\frac{Q}{\Delta T}= \frac{H_vm_f}{\Delta T}$

Here, C is the heat capacity and $\Delta T$ is the rise in temperature. We find the heat capacity of the calorimeter taking all these 4 substances. The subscript 1, 2, 3, 4 denotes the four values of heat capacities. Note that the values we obtain should not differ by much. Then we will take the average of this 4 values.

Wood (1)

Putting in the values, we have,

$C_1=\frac{H_vm_f}{\Delta T}=\frac{(18.9kJ/g)\times (1g)}{1.9^oC}=9.95kJ/^oC$

Coal (2)

$C_2=\frac{H_vm_f}{\Delta T}=\frac{(34.6kJ/g)\times (1g)}{3.5^oC}=9.89kJ/^oC$

Gasoline (3)

$C_3=\frac{H_vm_f}{\Delta T}=\frac{(46.5kJ/g)\times (1g)}{4.7^oC}=9.89kJ/^oC$

Ethanol (4)

$C_4=\frac{H_vm_f}{\Delta T}=\frac{(29.7kJ/g)\times (1g)}{3.0^oC}=9.90kJ/^oC$

Average Value

We take the average of 4 values of heat capacity,

$C=\frac{9.95+9.89+9.89+9.90}{4}kJ/^oC=9.91kJ/^oC$

Thus, the heat capacity up to 2 decimal places is 9.91 kJ/^oC