Question

Constant-volume calorimeters are sometimes calibrated by running a combustion reaction of known ΔE and measuring the change in temperature. For example, the combustion energy of glucose is 15.57 kJ/g. When a 1.500 g sample of glucose burns in a constant volume calorimeter, the calorimeter temperature increases from 21.45 to 23.34°C. Find the total heat capacity of the calorimeter (in kJ/K).

Answer 1

he total heat capacity of the calorimeter is - 0.09 kJ/K

Step-by-step explanation:

Given information

the combustion energy of glucose E = 15.57 kJ/g

mass of glucose, m = 1.500 g

initial temperature,
`T_(1)` = 21.45
`^(o) C`

final temperature
`T_(2)` = 23.34
`^(o) C`

the heat capacity of the calorimeter can be determined by

C = Q/ΔT

where

C = heat capacity (kJ/
`^(o) C`)

Q = quantity of the heat (kJ)

ΔT = temeperature difference (
`^(o) C`)

Q = E m

= - (15.57 kJ/g)(1.500 g)

= 23.355 kJ

ΔT =
`T_(2)`-
`T_(1)`

= 23.34
`^(o) C` - 21.45
`^(o) C`

= 1.89
`^(o) C`

Since the question ask to answer in kJ/k, we need to convert the temperature from celcius to kelvin by adding 273. thus,

ΔT = 1.89
`^(o) C` + 273

= 274.89 K

C = Q/ΔT

= - 23.355 kJ/274.89 K

= - 0.09 kJ/K