Calorimetry Formula


Calorimetry Formula

Calorimetry defines the act of measuring different changes in state variables of a body for deriving the heat transfer related to the changes of its states. This Calorimetry is performed with a calorimeter.

Formula for Calorimetry

Solved Examples

Problem 1: 4.409g sample of propane was burned along with excess oxygen in the bomb calorimeter. The temperature of the calorimeter raised by 6.85oC. Find how much energy of heat is released per mole of propane burned under these conditions.

Answer :

Molar mass of propane (C3H8) = 44.09g/mol.

This enables us to find out the number of moles of propane burned and therefore the amount of energy released. We need to use the heat capacity of the calorimeter to correct the effect of the calorimeter.

Moles C3H8 = 4.409g C3H8 ×× 1mol C3H844.09g C3H81mol C3H844.09g C3H8 = 0.1000 mol C3H8

Energy change per mole C3H8 = 32.4KJC32.4KJC ×× 6.85oC ×× 10.1000mol C3H810.1000mol C3H8

                                                  =  −2.22×103KJ C3H8mol C3H8−2.22×103KJ C3H8mol C3H8

We add a negative sign to the values because energy is released from this reaction, where the reaction is exothermic.

Problem 2: What quantity of ice at 0oC can be melted by 100J of heat?


Heat to fuse a substance = heat of fusion of the substance ×× mass of the substance.

This can be expressed by the following formula where q is used to denote heat measurement made in a calorimeter.

q = m (mass) ×× C (heat of fusion)

Solving for m we get

m = qCqC

    = 100J/3.34 ×× 102J/g

    = 29.9 ×× 10-2g or 0.299g of ice melted

Because heat is absorbed in this melting this is an endothermic reaction.


Leave a Comment

Your email address will not be published. Required fields are marked *