Physics
Cambridge IGCSE

TOPIC 2: THERMAL PHYSICS

# 2.2c Specific Heat Capacity

The Faroe Islands are a small island group in the far north of the Atlantic Ocean. They should be extremely cold in winter due to their far north location. However, the sea around the islands keeps them at a moderate temperature all year round. This is because water is a substance with a high specific heat capacity (S.H.C.) - a measure of how much energy is needed to change the temperature of a substance. The ocean water moving up from the south holds a huge reservoir of heat energy during winter, keeping the air warmer.

Figure 1: The Faroe Islands

Water has a higher S.H.C. than aluminium, because it takes longer on a cooker to heat 1kg of water than a 1kg aluminium sauce pan. More energy is required.
S.H.C. is defined as the energy required to change the temperature of an object by one degree Celsius per kilogram of mass. It is measured in joules per kilogram per degree Celsius (J/kg 0C).

From the definition above for specific heat capacity, we can obtain a formula:

 S.H.C. [J/kg 0C] = change in thermal energy [J] mass [kg] x temperature change [0C]
 c = ΔE m Δθ
Learn this formula!

Notes:

• The symbol for S.H.C. is just 'c'.
• Notice that E is the letter used for heat energy, and ΔE is the change in heat energy.
• Δθ is the increase (or decrease) in temperature.
As it is a change in temperature, it could be measured in 0C or kelvin, but 0C is fine.

The formula above can only be used for a substance in one state of matter. It does not work when a change of state is involved. Water has one of the highest S.H.C. values of household substances (4180 J/kg 0C) and this means it is difficult to heat it up. This is why 'cool packs' are often water based gels, as they stay cold for longer.

Example:
A kettle is used to boil some water. Calculate the heat energy required to heat 2 kg of water from 20 0C to boiling.

Rearranging the formula above gives ΔE = m c Δθ
We also know that m = 2 kg, c = 4180 J/kg 0C for liquid water, and Δθ = 80 0C (as water boils at 100 0C).
So ΔE = m c Δθ = 2 x 4180 x 80
ΔE = 668 800 joules or 669 kJ (to 3 sig. figs.)

### Required Practical: An investigation to determine the specific heat capacity of solids or liquids

You will need to be able to describe an investigation to find the S.H.C. of materials, usally a solid. Typically, aluminium or copper is used in a simple experiment:

You will need:

• A 1 kg block of copper or aluminium. (The block will have holes drilled for a heater and thermometer / temperature sensor).
• A thermometer or temperature sensor.
• A low voltage (e.g. 12 V) heating element.
• A 12 V power supply for the heating element.
• A joulemeter. (This measures the electrical energy transferred to the heating element. An ammeter and voltmeter could be used instead).
• Some insulating materials.

Method:

Set up the apparatus as shown in figure 2. Before the power pack is switched on, measure the temperature of the block if using a thermometer instead of a sensor and data logger. Then turn on the power supply.

The joulemeter will begin recording the energy in joules being transferred to the heating element, which we assume to be 100% efficient. When the temperature of the block has increased significantly, say 40 or 50 0C, turn off the power pack. Leave the apparatus for a short while as the temperature may increase further as the heat conducts through the metal. Then record the energy transferred and the temperature rise.

Figure 2: Investigating the S.H.C. of a metal block

For best results, insulate the block to prevent heat loss to the surroundings. (Not shown in the diagram). A drop or two of water in the temperature sensor hole will lead to better contact between the sensor/ thermometer and he block.

Results:

We can find the S.H.C. of the copper/aluminium by substituting our results into the rearranged formula above:

 c = ΔQ m x ΔT

### Liquids

To find the S.H.C. of water (or similar liquid), just use a known mass of water in a thin light beaker. Repeat the experiment above with the sensor and heating element in the water, and stir regularly to ensure the heat energy is distributed evenly. Some heat energy will transfer to the beaker, but if it has a low mass compared to the water it should not have a significant effect on the result.

Try some of the questions below to gain familiarity with the S.H.C. formula:

Questions:

1. A kettle is filled with 2.5 kg of water of S.H.C. 4180 J/kg 0C. Calculate the energy needed to heat the water from a room temperature of 25 0C to the boiling point at 100 0C .

We know that:
ΔE = m c Δθ
and so:
ΔE = 2.5 x 4180 x (100-25)
ΔE = 784 000 J
(784 kJ)

2. An investigation is carried out to find the S.H.C. of a 1 kg copper block. The block increases from 22 0C to 61 0C with 16 400 J of heat energy transferred.

• a) Show that the S.H.C. of copper is about 400 J/kg 0C
• b) The heating element has a power output of 48 W. Calculate the time it takes to heat the copper to 61 0C.

a) Δθ is 61-22 = 39 0C. Using the formula for S.H.C.:
 c = ΔE m x Δθ
So:
 c = 16 400 1 x 39

c= 421 J/kg 0C to 3 sig.figs., which is about 400 J/kg 0C.
(Do NOT just write 'about 400' here. Show the correct answer first).

b) we know from section 1.7b that power and energy /work are related by the formula:
 P = W t
Remember that work done is simply the energy transferred. Therefore:
 t = W P
 t = 16 400 48

t = 342 seconds

Extension

This question is very hard, but if you fancy a challenge, have a go and then check the solution below:

3. Ice cubes of mass 60 g are placed into a glass of 300 g of water. The ice is initially at -18 0C and warms up to -1 0C over a period of 4 minutes. Ice has a S.H.C. of 2110 J/kg 0C.

• a) Calculate the heat energy gained by the ice during this time.
• b) Calculate the temperature drop of the water as heat energy is transferred to the ice. The S.H.C. of water is 4180 J/kg 0C.
• C) Suggest one reason why the answer in (b) may be unrealistic and incorrect.

a) We know that:
ΔE = m c Δθ, and m= 60 g = 0.06 kg
Therefore for the ice,
ΔE = 0.06 x 2110 x 17
ΔE = 2150 J (to 3 sig. figs.)

b) The heat energy to warm the ice comes from the surrounding water. Therefore the water cools, losing the same quantity of heat energy.
As ΔE = m c Δθ , then using m = 0.3 kg gives:

 Δθ = ΔE m x c
 Δθ = 2150 0.3 x 4180
Δθ = 1.72 0C (to 3 sig. figs.)

c) The water may not cool by 1.71 0C due to any of the following:
Glass cools down as well / heat gain or loss due to surroundings / some ice may melt / ice may not cool by the same temperature drop across the cubes.

Tier

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 Extended Tier (Core and Supplementary content, Grades A* to G) Core Tier (Core content only, grades C to G) Remember my choice