Question

When 2.50 g of a certain hydrocarbon was completely combusted in a "bomb (constant-volume) calorimeter" with a heat capacity (excluding water) of 350 J/°C and which contained 2.00 liters of water (density = 1.00 g/mL and specific heat capacity = 4.184 J/°C•g), the resulting temperature change was measured to be 5.52°C. Calculate the thermal energy (in kJ) released per gram of hydrocarbon combusted. (1) 48.1 kJ/g (2) 0.773 kJ/g (3) 19.2 kJ/g (4) 18.5 kJ/g (5) 46.2 kJ/g

Answer 1

The thermal energy released per gram is 19.2 kJ/g.

(3) is correct option.

Step-by-step explanation:

Given that,

Weight of hydrocarbon = 2.50 g

Heat capacity
`c = 350 J/^(\circ)C`

We need to calculate the thermal energy released

Using formula of thermal energy

Heat released =heat absorb by calorimeter+heat absorb by water

`Q=c\Delta T+mc\Delta T`

Put the value into the formula

`Q=350*5.52+2000*4.184*5.52`

`Q=48123.36\ J`

Now, The thermal energy released per gram

`Q'=(Q)/(m)`

Put the value into the formula

`Q'=(48123.36)/(2.50)`

`Q'=19.2\ kJ/g`

Hence, The thermal energy released per gram is 19.2 kJ/g.