1.1.3 Energy changes in systems: specific heat capacity
Calculating a change in thermal energy
Definition
Change in thermal energy
The change in thermal energy is the change in the energy stored thermally by a system as its temperature changes, measured in joules, J\text{J}J.
- When an object's temperature rises, its thermal store increases, so energy must be transferred to it.
- When its temperature falls, energy is transferred away and its thermal store decreases.
- Calculate the change in thermal energy with ΔE=mcΔθ\Delta E = mc\Delta\thetaΔE=mcΔθ.
- Here ΔE\Delta EΔE is the change in thermal energy in J\text{J}J, mmm is the mass in kg\text{kg}kg, ccc is the specific heat capacity in J/kg∘C\text{J/kg}^\circ\text{C}J/kg∘C, and Δθ\Delta\thetaΔθ is the temperature change in ∘C^\circ\text{C}∘C.
- Find the temperature change from Δθ=final temperature−initial temperature\Delta\theta = \text{final temperature} - \text{initial temperature}Δθ=final temperature−initial temperature.
Example
A 2.0 kg2.0\ \text{kg}2.0 kg block with a specific heat capacity of 900 J/kg∘C900\ \text{J/kg}^\circ\text{C}900 J/kg∘C warms from 20 ∘C20\ ^\circ\text{C}20 ∘C to 35 ∘C35\ ^\circ\text{C}35 ∘C; find the increase in its thermal store.
- Write the equation: ΔE=mcΔθ\Delta E = mc\Delta\thetaΔE=mcΔθ.
- Find the temperature change: Δθ=35−20=15 ∘C\Delta\theta = 35 - 20 = 15\ ^\circ\text{C}Δθ=35−20=15 ∘C.
- Substitute the values: ΔE=2.0×900×15\Delta E = 2.0 \times 900 \times 15ΔE=2.0×900×15.
- Calculate: ΔE=27 000 J\Delta E = 27\,000\ \text{J}ΔE=27000 J, so the thermal store of the block increases by 27 000 J27\,000\ \text{J}27000 J.
Specific heat capacity
Definition
Specific heat capacity
The specific heat capacity of a substance is the amount of energy needed to raise the temperature of 1 kg1\ \text{kg}1 kg of the substance by 1 ∘C1\ ^\circ\text{C}1 ∘C.
- A substance with a high specific heat capacity needs more energy to produce a given temperature rise.
- A substance with a low specific heat capacity needs less energy for the same mass and temperature rise.
- So if equal masses of two materials receive the same energy, the one with the lower specific heat capacity heats up more.
- Rearrange the equation to find the specific heat capacity: c=ΔEmΔθc = \frac{\Delta E}{m\Delta\theta}c=mΔθΔE.
Common Mistake
- Do not confuse temperature with thermal energy: temperature is not the total energy stored, which also depends on mass and specific heat capacity.
- Always use the temperature change, not the final temperature, and convert the mass from grams to kilograms before substituting.
Practical
Investigation: the specific heat capacity of a metal block
You heat a metal block of known mass with an electric heater, measure the electrical work it does and the temperature rise, and use a graph of temperature against work done to find the specific heat capacity.
- Measure the mass of the metal block on a balance and record it in kg\text{kg}kg; the standard copper, iron or aluminium blocks are about 1 kg1\ \text{kg}1 kg and have two holes drilled in them.
- Wrap the block in insulation and stand it on a heatproof mat, so that little energy escapes to the surroundings.
- Slide a 12 V12\ \text{V}12 V immersion heater into the wider central hole of the block.
- Connect the heater in series with the power supply and an ammeter, and connect a voltmeter across the heater so you can measure its power.
- Use a pipette to drip a little water into the second hole, then push the thermometer in, so it makes good thermal contact with the block.
- Set the power supply to 12 V12\ \text{V}12 V, switch on, and record the ammeter and voltmeter readings, which should stay almost constant during the experiment.
- Read the starting temperature and start the stopclock at the same moment, then record the temperature every minute for ten minutes.
- Work out the heater power from P=IVP = IVP=IV, then the work done at each time from work done=P×t\text{work done} = P \times twork done=P×t, and complete your table.
- Plot temperature in ∘C^\circ\text{C}∘C on the vertical axis against work done in J\text{J}J on the horizontal axis, and draw a line of best fit through the straight part, as the very start of the graph may curve.
- Measure the gradient of the straight part, gradient=change in temperaturechange in work done\text{gradient} = \frac{\text{change in temperature}}{\text{change in work done}}gradient=change in work donechange in temperature.
- Find the specific heat capacity from c=1mass×gradientc = \frac{1}{\text{mass} \times \text{gradient}}c=mass×gradient1, because the gradient of this graph equals 1mc\frac{1}{mc}mc1.
- Only part of the electrical energy reaches the block's thermal store, since some warms the heater, the thermometer and the surroundings, so the measured value usually comes out a little too high; the insulation reduces this.
- The heater and block become very hot, so let them cool before handling them.
Exam technique
- In a calculation, show the equation, work out Δθ\Delta\thetaΔθ, substitute with the mass in kg\text{kg}kg, and give the answer with its unit.
- For the practical, find the heater power from P=IVP = IVP=IV and the energy transferred from work done=P×t\text{work done} = P \times twork done=P×t.
- Explain that only part of that energy raises the block's thermal store, so the block is insulated and the measured specific heat capacity is slightly too high.
Self review
- Define specific heat capacity.
- Write the equation ΔE=mcΔθ\Delta E = mc\Delta\thetaΔE=mcΔθ and give the unit of each quantity.
- In the investigation, how do you find the energy the heater transfers to the block?
- Two equal masses receive the same energy; which heats up more, the one with higher or lower specific heat capacity?
- In the investigation, why is the block insulated, and why is the measured specific heat capacity usually a little too high?
