Question

1. A 10.0 g sample of propane, C3H8, was combusted in a constant-volume bomb

calorimeter. The total heat capacity of the bomb calorimeter and water was 8.0
kJ/°C. The molar heat of combustion of propane is -2 222 KJ/mol. If the starting
temperature of the water was 20 °C, what will be the final temperature of the
bomb calorimeter?

Answer 1

The final temperature of the bomb calorimeter is 82.98°C

Step-by-step explanation:

The given information are;

The mass of the propane sample = 10.0 g

The heat capacity of the bomb calorimeter = 8.0 kJ/°C

The molar heat of combustion of propane is -2,222 kJ/mol.

The starting (initial) temperature of the water, T₁ = 20°C

The final temperature of the bomb calorimeter = T₂

The molar mas of propane = 44.1 g/mol

The number of moles, n, of propane present in 10.0 g of propane is found as follows;

`Number \ of \ moles \ of \ propane, \ n = (Mass \ of \ propane)/(Molar mass \ of \ propane) = (10)/(44.1) = 0.227 \ moles`Which gives;

The heat, Δh, released from the combustion of 10.0 g of propane = 0.227 × -2222 kJ/mol

Δh = -503.85 kJ/mol

Heat gained by the calorimeter = Heat released from the combustion of 10.0 g of propane = 503.85 kJ/mol

Change in heat, Δh
`_c` in the calorimeter = Heat capacity × Temperature change

Δh
`_c` = m × C

503.85 kJ/mol = 8.0 kJ/°C × (T₂ - 20°C)

T₂ = -503.85 kJ/mol/(8.0 kJ/°C) + 20°C = 82.98°C

The final temperature of the bomb calorimeter = 82.98°C