Question

A 0.74-g sample of a compound is burned in a bomb calorimeter, producing a temperature change from 23.02 oC to 27.65 oC. The heat capacity of the calorimeter is determined to be 4.78 kJ/oC. What is ΔE (aka ΔU, in kJ/g) for the combustion of this compound? Enter your answer as an integer.

Answer 1

It will be 29.90 KJ/gram

Step-by-step explanation:

We have given mass of compound burn m = 0.74 gram

Temperature is changes from
`23.02^(\circ)C\ to\ 27.65^(\circ)C`

So change in temperature
`\Delta T=27.65-23.02=4.63^(\circ)C`

Heat capacity is given as
`=4.78KJ/^(\circ)C`

We know that
`\Delta E` is given as

`\Delta E=Heat\ capacity* \Delta T=4.78* 4.63=22.13kj`

In KJ/gram it will be
`=(22.13)/(0.74)=29.90KJ/gram`