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Calorimetry Formula

The thermochemists use information about the specific heat capacity to determine the amount of energy that is either gained or released by reactions.

By trapping and measuring the energy as heat given off or absorbed the thermochemist can obtain information about energy transfers. This procedure is known as calorimetry and the apparatus in which it is performed is called a calorimeter.

Within the perfect Styrofoam calorimeter the energy as heat lost by the system equals, that gained by the surroundings. Mathematically Calorimetry Formula is represented by

The final temperature of the system and that of the surroundings within the calorimeter are identical.

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Calorimetry Problems

Solved problems based on calorimetry are given below.

Solved Examples

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

The molar mass of propane (C3H8) is 44.09g/mol. Knowing this it enables us to calculate 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 $\times$ $\frac{1mol\ C_{3}H_{8}}{44.09g\ C_{3}H_{8}}$ = 0.1000 mol C3H8

Energy change per mole C3H8 = $\frac{32.4KJ}{C}$ $\times$ 6.85oC $\times$ $\frac{1}{0.1000mol\ C_{3}H_{8}}$

$\frac{-2.22 \times 10^{3}KJ\ C_{3}H_{8}}{mol\ C_{3}H_{8}}$

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

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

Heat to fuse a substance = heat of fusion of the substance $\times$ 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) $\times$ C (heat of fuson)

Solving for m we get

m = $\frac{q}{C}$
= 100J/3.34 $\times$ 102J/g
= 29.9 $\times$ 10-2g or 0.299g of ice melted

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

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