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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.

Respuesta :

Answer:

It will be 29.90 KJ/gram

Explanation:

We have given mass of compound burn m = 0.74 gram

Temperature is changes from [tex]23.02^{\circ}C\ to\ 27.65^{\circ}C[/tex]

So change in temperature [tex]\Delta T=27.65-23.02=4.63^{\circ}C[/tex]

Heat capacity is given as [tex]=4.78KJ/^{\circ}C[/tex]

We know that [tex]\Delta E[/tex] is given as

[tex]\Delta E=Heat\ capacity\times \Delta T=4.78\times 4.63=22.13kj[/tex]

In KJ/gram it will be [tex]=\frac{22.13}{0.74}=29.90KJ/gram[/tex]

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