# Specific Heat Capacity ## Defining Specific Heat Capacity - The specific heat capacity of a substance is a measure of how much energy is needed to change its temperature - It is defined as:

The amount of thermal energy required to raise the temperature of 1 kg of a substance by 1 °C

- This quantity determines how resistant a material is to temperature change - Measured in units of **Joules per kilogram per Kelvin** ($\text{J kg}^{-1} \text{K}^{-1}$) or **Joules per kilogram per degree Celsius** ($\text{J kg}^{-1} \text{°C}^{-1}$) - Has the symbol $c$ - Different substances have vastly different specific heat capacities - Specific heat capacity is primarily used when considering liquids and solids ## What Specific Heat Capacity Tells Us - From the definition, it follows that: - The **greater the mass** of the material, the more thermal energy required to raise its temperature - The **greater the change in temperature**, the higher the thermal energy required to achieve this change ## Low vs High Specific Heat Capacity **Materials with LOW specific heat capacity:** - Heat up quickly (require less energy for a given temperature rise) - Cool down quickly (release less energy for a given temperature drop) - Examples include metals like copper and lead **Materials with HIGH specific heat capacity:** - Heat up slowly (require more energy for a given temperature rise) - Cool down slowly (release more energy for a given temperature drop) - Example: water has a very high specific heat capacity | **Substance** | **Specific heat capacity / $\text{J kg}^{-1} \text{K}^{-1}$** | |---------------|--------------------------------------------------------------| | Aluminium | 910 | | Copper | 390 | | Lead | 126 | | Glass | 500 - 680 | | Water | 4200 | | Mercury | 140 | The chart below illustrates the dramatic differences in specific heat capacity between common materials:

- Good electrical conductors, such as copper and lead, have low specific heat capacities - This makes them excellent conductors of heat as well - This is due to their free electrons which can transfer energy efficiently

Water's high specific heat capacity is why it's excellent for cooling systems and why coastal areas have more moderate climates

### Visualizing the Effect of Specific Heat Capacity - When equal masses of different materials receive the same power input, their temperatures rise at different rates - The chart below shows how 1 kg samples of copper and water heat up when supplied with 1000 W of power:

- Copper (low specific heat capacity) heats up much faster than water (high specific heat capacity) - This demonstrates why materials with low specific heat capacity are more responsive to temperature changes ## The Specific Heat Capacity Equation - The amount of thermal energy $E$ needed to raise the temperature by $\Delta\theta$ for a mass $m$ with specific heat capacity $c$ is: $$E = mc\Delta\theta$$ Where: - $E$ = change in thermal energy (J) - $m$ = mass of the substance (kg) - $c$ = specific heat capacity of the substance ($\text{J kg}^{-1} \text{K}^{-1}$ or $\text{J kg}^{-1} \text{°C}^{-1}$) - $\Delta\theta$ = change in temperature (K or °C)

A temperature change of 1 K is equal to a temperature change of 1 °C, so you don't need to convert between scales when calculating Δθ

## Worked Example **Question:** A kettle has a power rating of 1.7 kW. A mass of 650 g of liquid at 25 °C is poured into the kettle. When the kettle is switched on, it takes 3.5 minutes to start boiling. Calculate the specific heat capacity of the liquid. **Answer:** **Step 1: State the known quantities** - Power = 1.7 kW = $1.7 \times 10^3$ W - Time = 3.5 minutes = $3.5 \times 60$ s = 210 s - Mass, $m$ = 650 g = 0.65 kg - Temperature change, $\Delta\theta = 100 - 25 = 75$ °C **Step 2: Calculate the energy supplied** Using Energy = Power × Time: $$E = 1.7 \times 10^3 \times 210 = 3.57 \times 10^5 \text{ J}$$ **Step 3: State the thermal energy equation** $$E = mc\Delta\theta$$ **Step 4: Rearrange for specific heat capacity** $$c = \frac{E}{m\Delta\theta}$$ **Step 5: Substitute values and calculate** $$c = \frac{3.57 \times 10^5}{0.65 \times 75} = 7300 \text{ J kg}^{-1} \text{°C}^{-1}$$ ## Experimental Determination of Specific Heat Capacity - The general equation used is: $$\text{SPECIFIC HEAT CAPACITY} = \frac{\text{HEAT SUPPLIED TO SUBSTANCE}}{\text{MASS} \times \text{CHANGE IN TEMPERATURE}}$$ ### Method for Solids **Equipment:** - A block of the substance (preferably 1 kg in mass) - A thermometer - An immersion heater - A power source - A voltmeter, ammeter and stop-clock **Procedure:** 1. Drill two holes in the metal block: - One for the immersion heater - One for the thermometer 2. Measure and record the initial temperature of the block 3. Connect the immersion heater to the power supply with: - An ammeter in series - A voltmeter in parallel 4. Turn on the power supply and start the stop-clock 5. Record the voltage and current readings periodically 6. After 5 minutes (300 seconds): - Switch off the power supply - Stop the stop-clock 7. Monitor the thermometer and record the highest temperature reached - This may occur a few minutes after the heater is switched off (as heat distributes through the block) **Analysis:** - Calculate the energy supplied using: $$\text{Energy} = \text{Current} \times \text{Voltage} \times \text{Time}$$ $$E = IVt$$ - Rearrange the specific heat capacity equation: $$c = \frac{E}{m\Delta\theta}$$ - Substitute your values to find $c$ ### Method for Liquids **Equipment:** - A beaker of liquid (ideally containing 400 ml) - A thermometer - An immersion heater - A power source - A voltmeter, ammeter and stop-clock - A digital balance **Procedure:** 1. Measure and record the mass of the liquid using the balance 2. Place the immersion heater and thermometer in the liquid 3. Measure and record the initial temperature 4. Connect the heater to the power supply with: - An ammeter in series (with the heater) - A voltmeter in parallel (across the heater) 5. Turn on the power supply and start the stop-clock 6. Record voltage and current readings periodically 7. After 10 minutes (600 seconds): - Switch off the power supply - Stop the stop-clock 8. Monitor the thermometer and record the highest temperature reached - This may occur a few minutes after switching off **Analysis:** - Calculate the energy supplied: $$E = IVt$$ - Calculate the temperature change: $$\Delta\theta = T_{\text{final}} - T_{\text{initial}}$$ - Calculate specific heat capacity: $$c = \frac{E}{m\Delta\theta}$$ ### Evaluation of Both Methods **Sources of error:** - **Heat loss to surroundings** - Not all heat supplied goes into the substance - Some escapes to the air, bench, or equipment - This makes the calculated energy value too large - Results in a value for specific heat capacity that is too high - **Instrument precision** - Fluctuations in the power supply affect voltage and current - **Solution:** Take several periodic measurements and calculate an average - **For liquids:** Water may evaporate, reducing the effective mass

Always allow the thermometer reading to stabilize after switching off the heater before recording the final temperature

## Practical Applications - The specific heat capacity of different materials determines their suitability for various purposes: - **Cooking equipment:** Materials with low specific heat capacity (e.g., copper pans) heat up quickly - **Thermal storage:** Materials with high specific heat capacity (e.g., water in radiators) store and release heat slowly - **Heat sinks:** Aluminium is often used due to its moderate specific heat capacity and good thermal conductivity