Specific Heat Capacity
This topic covers how adding or removing thermal energy changes an object's temperature. It explains that this change depends on the object's mass and the material it is made from, quantified by its specific heat capacity.
Part of the ESAT Physics syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- Transferring thermal energy to an object increases its internal energy and usually its temperature, unless it's changing state.
- The temperature change for a given amount of energy depends on the object's mass and the substance it's made of.
- A substance with a high specific heat capacity (like water) requires a lot of energy to heat up by one degree.
- A substance with a low specific heat capacity (like copper) heats up very quickly with the same amount of energy.
- The change in temperature (ΔT) is the final temperature minus the initial temperature. A temperature decrease results in a negative ΔT, meaning energy has been lost.
- Heat capacity is a property of an entire object (in J/°C), while specific heat capacity is a property of a substance per unit mass (in J kg⁻¹ °C⁻¹).
Diagram
› Why does this happen?
Why different materials heat up at different rates
Temperature is a measure of the average kinetic energy of the particles (atoms or molecules) in a substance. When you heat something, you add thermal energy, which increases its internal energy.
In a simple material like a solid metal, most of this added energy makes the atoms vibrate faster. This directly increases their kinetic energy, so the temperature rises quickly. This means metals have a low specific heat capacity.
However, in a substance like water, the added energy is shared. Some of it increases the kinetic energy of the molecules, which raises the temperature. But a lot of the energy is also stored as potential energy in the forces between the molecules.
Because the added energy is shared between these two 'stores' (kinetic and potential), you have to add a huge amount of energy to get the same increase in temperature compared to a metal. This is why water has a very high specific heat capacity.
Formulae
ΔQ = m c ΔT Use this to find the thermal energy (ΔQ) transferred when an object of mass (m) and specific heat capacity (c) undergoes a temperature change (ΔT) without changing its physical state (e.g., from solid to liquid).
Definitions
- Specific Heat Capacity (c)
- The amount of thermal energy required to raise the temperature of 1 kilogram of a substance by 1 degree Celsius. Its unit is J kg⁻¹ °C⁻¹.
Worked example
A 400 g block of aluminium is heated by an electric heater with a power output of 200 W for 36 seconds. If the initial temperature of the block was 15 °C, what is its final temperature? The specific heat capacity of aluminium is 900 J kg⁻¹ °C⁻¹.
- 1
First, calculate the total thermal energy (ΔQ) supplied by the heater.
Energy = Power × timeΔQ = 200 W × 36 s = 7200 J - 2
Convert the mass from grams to kilograms.
m = 400 g = 0.4 kg - 3
Rearrange the specific heat capacity formula to find the temperature change (ΔT):
ΔT = ΔQ / (m c) - 4
Substitute the known values into the rearranged formula:
ΔT = 7200 / (0.4 × 900) - 5
Calculate the denominator:
0.4 × 900 = 4 × 90 = 360 - 6
Calculate the temperature change:
ΔT = 7200 / 360 = 720 / 36 = 20 °C - 7
Calculate the final temperature:
Final T = Initial T + ΔT = 15 °C + 20 °C = 35 °C
Answer: 35 °C
Common mistakes
- ×Forgetting to convert mass from grams to kilograms before substituting into the formula. The standard unit for 'c' uses kg, so mass must also be in kg.
- ×Making arithmetic errors, especially with decimals or large numbers. In the example, miscalculating 0.4 x 900 is a common source of error.
- ×Confusing energy (Joules) with power (Watts). You may need to first calculate energy using E = P × t if power and time are given.
No-calculator tips
- ✓Simplify calculations before you multiply. For `ΔT = 7200 / (0.4 × 900)`, you can write it as `7200 / 360`. Cancel a zero from top and bottom to get `720 / 36`. Since you know 36 x 2 = 72, then 36 x 20 = 720. So, the answer is 20.
- ✓Handle decimals by multiplying them out first. `0.4 × 900` is easier to think of as `4 × 90`, which is 360.
- ✓When rearranging the formula, do it algebraically before substituting numbers. This reduces the chance of calculation errors and keeps the process clear.