# Learn It: Sample Problems - Specific Heat Capacities

##### Exam tip:

When solving problems involving specific heat capacities, you only need to know two equations!

**1. Heat = mass × specific heat capacity × temperature change**

Q = mc Dq

and

**2. Heat lost = Heat gained**

mc Dq = mc Dq

When reading the question, concentrate on what is losing heat and what is gaining heat - then use the above two equations to solve the problem.

##### Worked Examples

**Question 1**

Given that the specific heat capacity of water is 11 times that of copper, calculate the mass of copper at a temperature of 100 °C required to raise the temperature of 200 g of water from 20.0 °C to 24.0 °C, assuming no energy is lost to the surroundings.

**Solution**

Heat lost by copper = heat gained by water

m_{cu}c_{cu}Dq_{cu} = m_{w}c_{w}Dq_{w}

m_{cu}c_{cu}(100 - 24) = 0.200 × 11c_{cu}(24 – 20)

76m_{cu} = 8.8

m_{cu} = 0.116 kg

**Question 2**

Three litres of water at 100 °C are added to 15 litres of water at 40 °C. Calculate the temperature of the mixture. Take the mass of 1 litre of water to be 1 kg and the specific heat capacity of water to be 4.2 × 103 J kg ^{-1} K ^{-1}

**Solution**

Let the temperature of the mixture q

Heat lost = heat gained

m_{1}c_{1}Dq_{1} = m_{2} c_{2} Dq_{2}

3 × 4.2 × 103 × (100 – q) = 15 × 4.2 × 103 × (q -40)

Solving for q gives q = 50°C

You try it!

**Question 3**

1 kg of water at a temperature of 45 °C is mixed with 1.5 kg of alcohol at 20 °C. Find the final temperature of the mixture.

Take the specific heat capacity of water to be 4200 J kg ^{–1} K ^{–1} and the specific heat capacity of alcohol to be 2400 J kg ^{–1} K ^{–1} . Assume no other exchange of heat occurs.

**Solution**

Let the final temperature of the mixture be q.

Heat lost by water = heat gained by alcohol

m_{w}c_{w}Dq_{w} = m_{al}c_{al}Dq_{al}

1 × 4200 × (45 – q) = 1.5 × 2400 × (q – 20)

(mcDq)_{w} = (mcDq)_{al}

1 × 4200 × (45 – q) = 1.5 × 2400 × (q – 20)

Solving for q gives q = 33 °C.

You try it!

Application of specific heat capacity: storage heaters

Some electrical heaters contain oil which has a high boiling point and a relatively high specific heat capacity. Once the oil has been heated it retains the thermal energy for quite a while, cooling down only slowly.

In storage heaters, a core of high-density thermal blocks (like very hard building blocks), is heated by electric elements during the night. These blocks are preferred to water, despite having a slightly lower specific heat capacity, because they are more dense (and thus more compact) and do not leak. They are heated during the night using low-cost (off-peak) electricity and release the stored thermal energy during the day.

Storage heaters maintain a steady room air temperature over long periods, warming the entire building fabric and reducing the risk of condensation.

Next: Learn It - Experiment 1

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