Specific Heat Capacity Equation

The Thermal Energy Equation

Every time you boil a kettle, a precise mathematical relationship links the energy transferred to the heat gained by the water. You can calculate the exact amount of energy transferred using the equation provided on the Edexcel physics equation sheet:

$\Delta Q = m \times c \times \Delta \theta$

In this equation, the Greek letter delta ($\Delta$) signifies a "change in", while theta ($\theta$) is the symbol Edexcel officially uses to represent temperature:

• $\Delta Q$ = change in thermal energy, measured in Joules (J)
• $m$ = mass, measured in Kilograms (kg)
• $c$ = specific heat capacity, measured in Joules per kilogram per degree Celsius (J/kg °C)
• $\Delta \theta$ = change in temperature, measured in degrees Celsius (°C)

It is highly useful to know that a temperature change of 1 °C is exactly equal to a temperature change of 1 Kelvin (K). If an exam question provides the temperature change in Kelvin, no numerical conversion is required.

Calculating and Rearranging the Formula

To solve for variables other than energy, you must be able to confidently rearrange the formula:

• To find specific heat capacity: $c = \frac{\Delta Q}{m \times \Delta \theta}$
• To find mass: $m = \frac{\Delta Q}{c \times \Delta \theta}$
• To find temperature change: $\Delta \theta = \frac{\Delta Q}{m \times c}$

Example 1: Calculating Energy ($\Delta Q$) Calculate the energy needed to heat 800 g of water from 20 °C to 45 °C ($c = 4200\text{ J/kg °C}$).

1. Convert Mass: $800\text{ g} \div 1000 = 0.8\text{ kg}$
2. Calculate $\Delta \theta$: $45\text{ °C} - 20\text{ °C} = 25\text{ °C}$
3. Substitute into formula: $\Delta Q = 0.8 \times 4200 \times 25$
4. Final Answer: $84,000\text{ J}$ (or $84\text{ kJ}$)

Example 2: Calculating Specific Heat Capacity ($c$) A 0.5 kg copper block absorbs 1520 J of thermal energy, rising in temperature by 8.0 °C. Calculate the specific heat capacity of copper.

1. State rearranged formula: $c = \frac{\Delta Q}{m \times \Delta \theta}$
2. Substitute values: $c = \frac{1520}{0.5 \times 8.0} = \frac{1520}{4}$
3. Final Answer: $380\text{ J/kg °C}$

Example 3: Calculating Mass ($m$) A hot water bottle releases 756,000 J of energy as it cools from 80 °C to 20 °C ($c = 4200\text{ J/kg °C}$). Calculate the mass of the water.

1. Calculate $\Delta \theta$: $80\text{ °C} - 20\text{ °C} = 60\text{ °C}$
2. State rearranged formula: $m = \frac{\Delta Q}{c \times \Delta \theta}$
3. Substitute and solve: $m = \frac{756,000}{4200 \times 60}$
4. Final Answer: $3\text{ kg}$

Level 9: Finding Final Temperature A 2 kg steel block ($c = 450\text{ J/kg °C}$) at 18 °C is supplied with 18,000 J. Find the final temperature.

1. Rearrange for $\Delta \theta$: $\Delta \theta = \frac{18,000}{2 \times 450} = 20\text{ °C}$
2. Calculate Final Temp: $18\text{ °C (initial)} + 20\text{ °C (change)} = 38\text{ °C}$

Mathematical Skills and Exam Traps

The most common pitfall is the "grams trap". Mass is often provided in grams, but it must be converted to kilograms (by dividing by 1,000) before substituting it into the equation. Similarly, thermal energy might be given in kilojoules (kJ) or megajoules (MJ), requiring you to multiply by 1,000 or 1,000,000 respectively to reach Joules.

When dealing with cooling substances, the temperature change is technically negative, resulting in a negative energy value which indicates energy released. For mass or specific heat capacity calculations, use the absolute (positive) value of the temperature change.

Finally, always check if the question asks for the final temperature rather than the temperature change. If asked for final temperature, you must calculate $\Delta \theta$ first, then add it to the initial starting temperature. Round your final answer to the same number of significant figures as the least precise value provided in the question.