P4: Specific Heat Capacity
Specific heat capacity and calculations — the SHC equation, the required practical, and why water is used in heating systems.
P4 Specific Heat Capacity
📋 Key Definitions
Specific heat capacity (c): The amount of energy needed to change the temperature of 1 kg of a substance by 1 °C. Measured in joules per kilogram per degree Celsius (J/kg°C).
Key idea: Different substances need different amounts of energy to change their temperature by the same amount. A substance with a high SHC takes a lot of energy to heat up, but also stores a lot of energy.
📐 The SHC Equation
change in thermal energy = mass x specific heat capacity x change in temperature
ΔE = m x c x ΔT
Where:
- ΔE = change in thermal energy, in joules (J)
- m = mass, in kilograms (kg)
- c = specific heat capacity, in J/kg°C
- ΔT = change in temperature, in °C
Rearranging the equation: c = ΔE / (m x ΔT). m = ΔE / (c x ΔT). ΔT = ΔE / (m x c).
📊 SHC Values for Common Substances
| Substance | Specific Heat Capacity (J/kg°C) | What this means |
|---|---|---|
| Water | 4200 | Needs a lot of energy to heat up — stores lots of energy |
| Aluminium | 900 | Heats up quickly with less energy |
| Iron | 450 | Heats up even more quickly |
| Copper | 385 | Heats up quickly — good for saucepans |
| Lead | 130 | Very easy to change temperature |
| Oil | ~2000 | Less than water, heats up faster than water |
Water has a very high SHC — it takes 4200 J to heat 1 kg of water by 1 °C. This means water stores a lot of thermal energy, which is why it is used in central heating systems and as a coolant.
🧪 Required Practical: Measuring Specific Heat Capacity
Aim: To measure the specific heat capacity of a material (typically a metal block).
Method
- Measure the mass of the metal block using a balance.
- Wrap the block in insulation to reduce heat loss to the surroundings.
- Insert an immersion heater and a thermometer into holes in the block.
- Connect the immersion heater to a joulemeter (or to a voltmeter, ammeter and stopwatch to calculate E = V x I x t).
- Record the initial temperature of the block.
- Switch on the heater and start the stopwatch. Leave it on for a set time (e.g. 10 minutes).
- Record the final temperature of the block and the energy supplied by the heater.
- Calculate SHC: c = E / (m x ΔT)
Sources of Error
- Heat loss to surroundings — not all energy goes into the block, so calculated SHC is too high
- Thermometer not in good thermal contact — may not read true temperature
- Heat not distributed evenly — the block may have hot spots near the heater
Improving the Practical
Use thicker insulation around the block, use a lid to reduce convection, stir if using a liquid, and allow time for heat to distribute evenly before measuring the final temperature.
💧 Why Water Is Used in Heating Systems
Water is ideal for central heating systems because its high specific heat capacity means it can store a large amount of thermal energy and release it slowly as it flows through radiators. This makes it an effective heat transfer fluid.
- High SHC (4200 J/kg°C) — stores lots of energy per kg
- Can carry heat from boiler to radiators efficiently
- Releases heat slowly, keeping rooms warm for longer
- Also used as a coolant in car engines for the same reason
📝 Worked Examples
Worked Example 1 — Calculating Energy Needed
Question: Calculate the energy needed to heat 2 kg of water from 20 °C to 80 °C. The specific heat capacity of water is 4200 J/kg°C.
Solution:
ΔE = m x c x ΔT = 2 x 4200 x (80 - 20) = 2 x 4200 x 60 = 504 000 J = 504 kJ
Worked Example 2 — Calculating SHC
Question: A 0.5 kg aluminium block is heated with 18 000 J of energy. Its temperature rises from 20 °C to 60 °C. Calculate the specific heat capacity of aluminium.
Solution:
c = ΔE / (m x ΔT) = 18 000 / (0.5 x 40) = 18 000 / 20 = 900 J/kg°C
Worked Example 3 — Calculating Temperature Change
Question: 200 000 J of energy is transferred to 10 kg of water (c = 4200 J/kg°C). Calculate the temperature rise.
Solution:
ΔT = ΔE / (m x c) = 200 000 / (10 x 4200) = 200 000 / 42 000 = 4.76 °C
The temperature rises by approximately 4.8 °C.
Worked Example 4 — Required Practical Calculation
Question: In a SHC experiment, an immersion heater is connected to a 12 V supply and carries a current of 4 A for 5 minutes. The 0.8 kg metal block rises in temperature from 22 °C to 72 °C. Calculate the SHC.
Solution:
Energy supplied: E = V x I x t = 12 x 4 x (5 x 60) = 12 x 4 x 300 = 14 400 J
c = E / (m x ΔT) = 14 400 / (0.8 x 50) = 14 400 / 40 = 360 J/kg°C
Worked Example 5 — Comparing Substances
Question: Equal amounts of energy (10 000 J) are supplied to 1 kg of water (c = 4200) and 1 kg of copper (c = 385). Which has the larger temperature rise?
Solution:
Water: ΔT = 10 000 / (1 x 4200) = 2.38 °C
Copper: ΔT = 10 000 / (1 x 385) = 25.97 °C
Copper has a much larger temperature rise because it has a much lower SHC — it takes less energy to change its temperature.
Worked Example 6 — Real-World Application
Question: A kettle contains 0.5 kg of water at 25 °C. The kettle has a power rating of 2000 W. How long will it take to boil the water (100 °C)? c = 4200 J/kg°C.
Solution:
Energy needed: ΔE = 0.5 x 4200 x (100 - 25) = 0.5 x 4200 x 75 = 157 500 J
Time = ΔE / P = 157 500 / 2000 = 78.75 s (approximately 79 seconds)
❓ Practice Questions
Q1: Foundation Define specific heat capacity and state its unit.
Q2: Foundation Calculate the energy needed to heat 3 kg of water from 15 °C to 85 °C. c = 4200 J/kg°C.
Q3: Foundation Explain why water is used in central heating systems rather than oil.
Q4: Higher A 1.5 kg block of iron (c = 450 J/kg°C) is heated from 20 °C to 120 °C. Calculate the energy supplied.
Q5: Higher In a required practical, a 0.4 kg metal block is heated by a 10 V, 2 A heater for 7 minutes. The temperature rises by 35 °C. Calculate the SHC. Explain why the true SHC may be different from your calculated value.
✅ Answers
- The specific heat capacity of a substance is the amount of energy needed to change the temperature of 1 kg of the substance by 1 °C. The unit is J/kg°C (joules per kilogram per degree Celsius).
- ΔE = m x c x ΔT = 3 x 4200 x (85 - 15) = 3 x 4200 x 70 = 882 000 J = 882 kJ.
- Water has a higher SHC (4200 J/kg°C) than oil (~2000 J/kg°C). This means water stores more thermal energy per kg for each degree of temperature rise. Water can therefore carry more energy from the boiler to the radiators and release it over a longer time, making it more effective for heating.
- ΔE = m x c x ΔT = 1.5 x 450 x (120 - 20) = 1.5 x 450 x 100 = 67 500 J = 67.5 kJ.
- E = V x I x t = 10 x 2 x (7 x 60) = 10 x 2 x 420 = 8400 J. c = E / (m x ΔT) = 8400 / (0.4 x 35) = 8400 / 14 = 600 J/kg°C. The true SHC may be different because some energy is lost to the surroundings (not all heat goes into the block), so the calculated SHC is higher than the true value. Also, the thermometer may not read the true average temperature of the block.
🎯 Exam Tips
- Always convert mass to kilograms before using the equation — a common mistake is using grams.
- ΔT is the temperature change, not the final temperature. Calculate ΔT = T_final minus T_initial.
- In the required practical, if using E = VIt, make sure you convert time to seconds.
- If your calculated SHC seems too high, it is likely because of heat loss to the surroundings — not all the energy goes into heating the block.
- When comparing substances: high SHC = hard to heat, stores lots of energy; low SHC = easy to heat, stores less energy.
- The water SHC value (4200 J/kg°C) comes up frequently — memorise it.
🔬 Required Practical
Required Practical: Measuring the Specific Heat Capacity of a Material
Aim: To determine the specific heat capacity of a metal block by measuring the energy supplied and the temperature change.
Method: 1) Measure the mass of the metal block using a balance. 2) Wrap the block in insulation and insert an immersion heater and thermometer. 3) Record the initial temperature. 4) Connect the heater to a joulemeter (or use a voltmeter, ammeter and stopwatch for E = V × I × t). 5) Switch on the heater for a set time (e.g. 10 minutes). 6) Record the final temperature and energy supplied. 7) Calculate c = E ÷ (m × ΔT).
Variables: IV: Energy supplied to the block (or time of heating), DV: Temperature change of the block, Control: Mass of block, type of material, insulation used
🔢 Maths Skills
Mathematical Skills
You need to use the equation ΔE = mcΔT and rearrange it to find any unknown. You must convert units correctly (grams to kilograms, minutes to seconds) and use standard form for large energy values. Calculating energy from electrical measurements using E = VIt also requires unit conversion.
Maths Example
An immersion heater runs on 12 V and carries 3.5 A of current for 8 minutes. Calculate the energy supplied, then find the SHC of a 1.2 kg block that rises in temperature by 28 °C.
Time = 8 × 60 = 480 s. Energy E = V × I × t = 12 × 3.5 × 480 = 20 160 J. c = E ÷ (m × ΔT) = 20 160 ÷ (1.2 × 28) = 20 160 ÷ 33.6 = 600 J/kg°C.
⚠️ Common Misconceptions
Watch Out!
1. Wrong: A high SHC means something gets very hot quickly Correct: A high SHC means a substance needs MORE energy to change its temperature — it heats up slowly and stores lots of thermal energy
2. Wrong: Temperature and thermal energy are the same thing Correct: Temperature measures how hot something is (°C), while thermal energy depends on mass, SHC and temperature change — a large bath of warm water has more thermal energy than a small cup of boiling water
3. Wrong: You can use mass in grams directly in the SHC equation Correct: Mass must always be in kilograms — 200 g must be converted to 0.2 kg before using ΔE = mcΔT
✍️ 6-Mark Question
Extended Answer
6 marks: Describe how you would investigate the specific heat capacity of a metal block in the laboratory. Explain your method and evaluate the main sources of error and how they affect the result.
Method: Measure the mass of the metal block using a balance. Wrap the block in insulation to reduce heat loss. Insert an immersion heater and a thermometer into the holes in the block. Record the initial temperature. Connect the immersion heater to a joulemeter (or use a voltmeter, ammeter and stopwatch and calculate E = V × I × t). Switch on the heater for a set time (e.g. 10 minutes). Record the final temperature and the total energy supplied. Calculate the SHC using c = E ÷ (m × ΔT). Sources of error: (1) Heat loss to the surroundings — not all the electrical energy goes into heating the block; some is dissipated to the thermal store of the air and insulation. This means the calculated SHC is higher than the true value because less energy actually heats the block than the joulemeter records. (2) The thermometer may not be in good thermal contact with the block, so the measured temperature change may be inaccurate. (3) Heat may not be distributed evenly through the block, creating hot spots near the heater. To improve: use thicker insulation, allow time for heat to distribute before reading the final temperature, and use a smaller temperature rise to reduce heat loss.
Mark scheme: 1 mark — correct method with mass, heater, thermometer and insulation; 1 mark — measuring energy and temperature change; 1 mark — correct use of equation c = E/(mΔT); 1 mark — heat loss to surroundings identified and effect on result explained; 1 mark — second error identified (thermometer contact or uneven heating); 1 mark — suggestion for improvement
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
A student heats two blocks with the same immersion heater for the same time. Block A is aluminium (mass 1.0 kg, c = 900 J/kg°C) and Block B is iron (mass 1.0 kg, c = 450 J/kg°C). The heater supplies 18 000 J of energy to each block. Both blocks start at 20 °C.
(a) Calculate the temperature rise and final temperature of each block.
(b) The student's results show that the aluminium block only rises by 16 °C instead of the calculated value. Explain why the actual temperature rise is less than expected.
(c) A student claims "Iron must store more thermal energy than aluminium because it reaches a higher temperature." Evaluate this claim.
Answers: (a) Block A: ΔT = 18 000 ÷ (1.0 × 900) = 20 °C, final temperature = 20 + 20 = 40 °C. Block B: ΔT = 18 000 ÷ (1.0 × 450) = 40 °C, final temperature = 20 + 40 = 60 °C. (b) The actual temperature rise is less because some energy is lost to the surroundings (thermal store of the air and insulation) — not all 18 000 J goes into heating the block. Also, heat may not be distributed evenly, so the thermometer may not read the true average temperature. (c) The claim is incorrect. The thermal energy stored in each block is the same (18 000 J was supplied to each). Iron reaches a higher temperature precisely BECAUSE it has a lower SHC — it needs less energy per kg per °C, so the same energy causes a bigger temperature rise. Aluminium stores the same amount of energy but at a lower temperature because its higher SHC means more energy is needed per degree. Temperature and thermal energy are not the same thing.
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