# Calculations involving Specific Heat Capacity (Engineering Science)

## Calculations involving Specific Heat Capacity (Engineering Science)

The specific heat capacity, c, of a substance is the amount of energy needed to raise the temperature of a 1kg of the substance by 1K ( or 1 °C) without change of state.

The unit of specific heat capacity is Jkg-1K-1 or Jkg-1°C-1, which is usually expressed as J/kgK or J/kg°C or J/(kgK) or J/(kg°C).

The table below shows specific heat capacities of some common substances.

Substance Specific heat capacity (J/kgK)
Aluminium 900
Concrete 850
Copper 390
Iron 490
Oil 2100
Water 4200
Glass 500–680
Ice 2100
Seawater 3950
Ethanol 2500
Mercury 140

### Calculating Specific Heat Capacity

To raise the temperature of an object of mass m from temperature θ1 to temperature θ2, the energy needed is given by the formula Q =mcΔθ, whereby:

• Q is the energy needed (in joules)
• m is the mass of the substance (in kg)
• c is the specific heat capacity (in J/kgK or J/kg°C)
• Δθ is the temperature rise (in K or °C) and Δθ = θ2 – θ1

Rearranging the equation gives us:

$$c = \frac{Q}{mΔθ}$$

### Example

26400J of energy is supplied to a 2kg block of aluminium and its temperature rises from 20°C to 35°C. Assuming that the block is well insulated so that there is no energy loss to the surroundings, determine the specific heat capacity of aluminium.

To calculate specific heat we use the formula $$c = \frac{Q}{mΔθ}$$

We are given the following values:

• Q = 26400J
• m = 2.0kg
• Δθ = (35−20)°C = 15°C

So we substitute these values into the equation and solve the equation:

$$c = \frac{Q}{mΔθ}$$

$$c = \frac{26400}{2 × 15}$$

$$= 880Jkg^{−1}K^{−1}$$

### Factors affecting the temperature rise of an object

The temperature rise of an object when it is heated depends on:

• mass of the object
• amount of energy supplied to it
• specific heat capacity of the object.

For objects of the same mass the higher the specific heat capacity, the more the energy needed to raise the temperature by the same degree.

For example, it takes more energy to raise the temperature of 1 kg of water by 1 °C than to raise the temperature of 1 kg of alcohol by the same amount. That is because water has a higher specific heat capacity than alcohol.

### Question 1 – Calculating the energy required to raise the temperature of a substance

Calculate the energy which must be supplied to raise the temperature of 5.0kg of water from 20°C to 100°C.

### Solution 1

First of all you do data collection:

• Temperature rise, Δθ = 100°C – 20°C = 80°C
• Mass of water, m = 5kg
• Specific heat capacity of water = 4 200 J/kg°C
• Energy supplied, Q = ?

Then you apply the given data to the formula Q =mcΔθ

• Q =mcΔθ
• Q = 5 × 4 200 × 80
• Q = 1 680 000 J
• Q = 1 680 kJ or 1,68 MJ

### Question 2 – Comparing the energy required to raise the temperature of different substances

Which requires more energy, heating a 2.0kg block of lead by 30K, or heating a 4.0kg block of copper by 5.0K?

### Solution 2

In this question we are comparing the energy required by lead to the energy required by copper. So we will collect the data for lead and the data for copper separately.

• Temperature rise, Δθ = 30 K
• mass = 2 kg
• Q = ?

So, heat energy required by lead,

• Q = mcΔθ
• Q = 2 × 130 × 30
• Q = 7 800 J

Data for copper:

• Temperature rise, Δθ = 5 K
• mass = 4 kg
• Specific heat capacity of lead, ccopper = 390 J/kg°C
• Q = ?

So, heat energy required by copper,

• Q = mcΔθ
• Q = 4 × 390 × 5
• Q = 7 800 J

Therefore, they require the same amount of heat energy.

### Question 3 – Calculating specific heat capacity of iron

A well-insulated 1.2kg block of iron is heated using a 50W heater for 4.0min. The temperature of the block rises from 22°C to 45°C. Find the experimental value for the specific heat capacity of iron.

### Solution 3

First of all you do data collection:

• Temperature rise, Δθ = 45°C – 22°C = 23°C
• Mass of iron, m = 1.2 kg
• Power of the heat, P = 50 W
• Time = 4 min = 240 s
• Specific heat capacity of iron, ciron = ?

For the formula to work, you first need to calculate the energy supplied by the heater. From the definition of power we can get the formula P = Q/t.

Therefore:

• Q = Pt
• Q = 50 × 240
• Q = 12 000 J

Now you apply the formulae Q = mcΔθ

• Q = mcΔθ
• 12 000 = 1.2 × ciron × 23
• 12 000 = 27.6ciron
• ciron = 12 000/27.6
• ciron = 435 J/kg°C

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