Answer to Question #271900 in Organic Chemistry for Piolo

Question

In a bomb calorimeter, 50g of a fuel was placed into the sample dish and was

allowed to undergo combustion. After the combustion, the water in the

calorimeter experienced an increase in temperature from 28°C to 32°C. Calculate

the heat of combustion of the fuel. Assume the volume of the calorimeter is 4 liters

and density changes in the water to be negligible. Use the heat capacity of water

to be 4.18 J/g°C. Assume that the calorimeter does not absorb any of the heat from

the combustion reaction.

Answer 1

A bomb calorimeter is a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction.

Volume of calorimeter = 4L

Heat capacity of calorimeter is negligible

Heat capacity of water = 4.18J/g°C

As given 50 g of fuel is given

The water in the calorimeter is warm up from 28°C to 32°C

Mass of water = (1000g/ml )×4L = 4000 g = 4kg

q combustion = - q water

= ms∆T = 4000×4.18×(4)

= - 66.88 kJ

50 g of fuel used

Heat of combustion = -66.88kJ/50g

= - 1.34kJ/g

2. 1.5 g of ethanol with 300 ml of water

Density of water = 1g/ml

Mass of water = 300×1 = 300 g

Water temperature increased by 55°C

∆T of water = 55°C

qethanol = - qwater

= - 300×4.18×55 = - 68.97 kJ

qethanol of 1.5 g = -68.97 kJ

Heat of combustion of ethanol = -68.97/1.5  

= 45.98 kJ/g