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

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

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

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

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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 6.85^{o}C. Calculate how much energy of heat is released per mole of propane burned under these conditions.

** Solution: **

The molar mass of propane (C_{3}H_{8}) 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 C_{3}H_{8} = 4.409g C_{3}H_{8} $\times$ $\frac{1mol\ C_{3}H_{8}}{44.09g\ C_{3}H_{8}}$ = 0.1000 mol C_{3}H_{8}

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

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

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 0^{o}C 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 $\times$ 10^{2}J/g

= 29.9 $\times$ 10^{-2}g or 0.299g of ice melted

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

The molar mass of propane (C

Moles C

Energy change per mole C

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

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

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 $\times$ 10

= 29.9 $\times$ 10

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