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Thermal Physics - Specific heat capacity

Grade 11IGCSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Specific heat capacity (cc) is defined as the amount of energy required to raise the temperature of 1 kg1 \text{ kg} of a substance by 1C1 ^\circ\text{C} (or 1 K1 \text{ K}).

The internal energy of a substance increases when it is heated, causing an increase in the average kinetic energy of its particles, which is observed as a temperature rise.

The SI unit for specific heat capacity is \text{J kg}^{-1} ^\circ\text{C}^{-1} or J kg1 K1\text{J kg}^{-1} \text{ K}^{-1}.

Substances with a high specific heat capacity, such as water (c \approx 4200 \text{ J kg}^{-1} ^\circ\text{C}^{-1}), require more energy to change their temperature compared to substances with low specific heat capacity like lead (c \approx 130 \text{ J kg}^{-1} ^\circ\text{C}^{-1}).

Thermal capacity (or heat capacity) is the energy required to raise the temperature of an entire object by 1C1 ^\circ\text{C}, calculated as C=mcC = mc.

In experimental setups, lagging (insulation) is used to minimize thermal energy loss to the surroundings to ensure accurate measurements of cc.

📐Formulae

Q=mcΔθQ = mc\Delta\theta

c=QmΔθc = \frac{Q}{m\Delta\theta}

C=mcC = mc

P×t=mcΔθP \times t = mc\Delta\theta

💡Examples

Problem 1:

A 2.0 kg2.0 \text{ kg} block of copper is heated from 25C25 ^\circ\text{C} to 75C75 ^\circ\text{C}. Given that the specific heat capacity of copper is 390 \text{ J kg}^{-1} ^\circ\text{C}^{-1}, calculate the thermal energy supplied.

Solution:

Q=mcΔθ=2.0×390×(7525)=2.0×390×50=39,000 JQ = mc\Delta\theta = 2.0 \times 390 \times (75 - 25) = 2.0 \times 390 \times 50 = 39,000 \text{ J}

Explanation:

To find the energy, we multiply the mass (m=2.0 kgm = 2.0 \text{ kg}), the specific heat capacity (c = 390 \text{ J kg}^{-1} ^\circ\text{C}^{-1}), and the change in temperature (Δθ=50C\Delta\theta = 50 ^\circ\text{C}).

Problem 2:

An electric immersion heater rated at 50 W50 \text{ W} is used to heat 0.5 kg0.5 \text{ kg} of a liquid for 4 minutes4 \text{ minutes}. The temperature of the liquid rises from 20C20 ^\circ\text{C} to 32C32 ^\circ\text{C}. Calculate the specific heat capacity of the liquid.

Solution:

Energy supplied Q=P×t=50×(4×60)=12,000 JQ = P \times t = 50 \times (4 \times 60) = 12,000 \text{ J}. Using c = \frac{Q}{m\Delta\theta} = \frac{12,000}{0.5 \times (32 - 20)} = \frac{12,000}{0.5 \times 12} = \frac{12,000}{6} = 2,000 \text{ J kg}^{-1} ^\circ\text{C}^{-1}

Explanation:

First, calculate the total energy supplied by the heater (P×tP \times t). Then, rearrange the specific heat capacity formula to solve for cc using the mass and the observed temperature change.

Specific heat capacity - Revision Notes & Key Formulas | IGCSE Grade 11 Physics