Question

A 1.0 g sample of propane, C3H8, was burned in calorimeter. The temperature rose from 28.5 0C to 32.0 0C and heat of combustion 10.5 kJ/g. Calculate the heat capacity of the calorimeter apparatus in kJ/0C

Answer 1

Answer:

A 1.0 g sample of propane, C3H8, was burned in the calorimeter.

The temperature rose from 28.5 0C to 32.0 0C and the heat of combustion 10.5 kJ/g.

Calculate the heat capacity of the calorimeter apparatus in kJ/0C

Explanation:

$Heat of combustion = heat capacity of calorimeter * deltaT$

Given,

The heat of combustion = 10.5kJ/g.

$deltaT = (32.0-28.5)^oC\\deltaT = 3.5^oC$

Substitute these values in the above formula to get the value of heat capacity of the calorimeter.

$deltaT =heat capacity of calorimeter * (change in temperature)\\10.5kJ/g = heat capacity of calorimeter * (3.5^oC)\\\\=>heat capacity of calorimeter = \frac{10.5kJ/g}{3.5^oC} \\=>heat capacity of calorimeter = 3.0 kJ/g.^oC$

Answer:

The heat capacity of the calorimeter is $3.0kJ/g.^oC.$