We’re being asked to calculate the heat capacity of the calorimeter. This is given by the equation:

$\overline{){{\mathbf{q}}}_{{\mathbf{rxn}}}{\mathbf{=}}{{\mathbf{q}}}_{{\mathbf{calorimeter}}}{\mathbf{+}}{{\mathbf{q}}}_{{\mathbf{solution}}}}$

where **q _{calorimeter} = heat absorbed by the calorimeter** and

Expanding this, we have:

${\mathbf{q}}_{\mathbf{rxn}}\mathbf{=}{\mathbf{C}}_{\mathbf{cal}}\mathbf{\u2206}\mathbf{T}\mathbf{+}\mathbf{mc}\mathbf{\u2206}\mathbf{T}$

where

C_{cal} = heat capacity of the calorimeter

m = mass of water (in grams)

c = specific heat capacity of water

ΔT = change in temperature = final T – initial T.

The combustion of 1 mole of glucose C_{6}H_{12}O_{6} releases 2.82×10^{3} kJ of heat. If 1.25 g of glucose are burnt in a calorimeter containing 0.95 kg of water and the temperature of the entire system raises from 20.10 °C to 23.25 °C. What is the heat capacity of the calorimeter?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Constant-Volume Calorimetry concept. You can view video lessons to learn Constant-Volume Calorimetry. Or if you need more Constant-Volume Calorimetry practice, you can also practice Constant-Volume Calorimetry practice problems.