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Physics - Heat Transfer (Conduction, Convection, Radiation, Thermal Expansion)

Grade 8ICSE

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

🔑Concepts

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Heat is a form of energy that flows from a body at a higher temperature to a body at a lower temperature. Its SI unit is the Joule (JJ), and another common unit is the calorie (calcal), where 1 cal≈4.186 J1 \text{ cal} \approx 4.186 \text{ J}.

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Conduction: The process of heat transfer in solids where energy is passed from one particle to another through vibrations without the actual movement of the particles. Metals are good conductors because they contain free electrons.

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Convection: The transfer of heat in fluids (liquids and gases) by the actual movement of the particles. This leads to the formation of convection currents. Examples include sea breeze and land breeze.

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Radiation: The transfer of heat in the form of electromagnetic waves (Infrared rays). It does not require a material medium and can travel through a vacuum. The speed of radiation is 3×108 m/s3 \times 10^8 \text{ m/s}.

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Thermal Expansion: Most substances expand when heated. The increase in length is called linear expansion, increase in area is superficial expansion, and increase in volume is cubical expansion.

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Anomalous Expansion of Water: Water exhibits unusual behavior; it contracts when heated from 0∘C0^\circ C to 4∘C4^\circ C. It has maximum density and minimum volume at 4∘C4^\circ C.

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Specific Heat Capacity: The amount of heat energy required to raise the temperature of a unit mass of a substance by 1∘C1^\circ C (or 1 K1\text{ K}).

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Thermos Flask: Also known as a Dewar flask, it is designed to minimize heat loss by conduction (using a vacuum between double walls), convection (sealed stopper), and radiation (silvered inner surfaces).

📐Formulae

Q=mcΔTQ = mc\Delta T

ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T

Lt=L0(1+αΔT)L_t = L_0 (1 + \alpha \Delta T), where LtL_t is length at temperature TT.

α=β2=γ3\alpha = \frac{\beta}{2} = \frac{\gamma}{3}

Heat Capacity (C′)=QΔT=mc\text{Heat Capacity } (C') = \frac{Q}{\Delta T} = mc

💡Examples

Problem 1:

A brass rod of length 2 m2\text{ m} is heated from 30∘C30^\circ C to 80∘C80^\circ C. If the coefficient of linear expansion of brass is α=1.9×10−5 ∘C−1\alpha = 1.9 \times 10^{-5} \text{ } ^\circ C^{-1}, calculate the increase in length.

Solution:

Given: L0=2 mL_0 = 2\text{ m}, T1=30∘CT_1 = 30^\circ C, T2=80∘CT_2 = 80^\circ C, α=1.9×10−5 ∘C−1\alpha = 1.9 \times 10^{-5} \text{ } ^\circ C^{-1}. Change in temperature ΔT=80−30=50∘C\Delta T = 80 - 30 = 50^\circ C. Using the formula: ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T ΔL=2×(1.9×10−5)×50\Delta L = 2 \times (1.9 \times 10^{-5}) \times 50 ΔL=100×1.9×10−5=1.9×10−3 m\Delta L = 100 \times 1.9 \times 10^{-5} = 1.9 \times 10^{-3} \text{ m} or 0.19 cm0.19 \text{ cm}.

Explanation:

The change in length is directly proportional to the original length and the change in temperature. By plugging the known values into the linear expansion formula, we find the expansion in meters.

Problem 2:

Calculate the amount of heat energy required to raise the temperature of 500 g500\text{ g} of copper from 20∘C20^\circ C to 70∘C70^\circ C. (Specific heat capacity of copper c=0.39 J g−1 ∘C−1c = 0.39 \text{ J g}^{-1} \text{ } ^\circ C^{-1})

Solution:

Given: m=500 gm = 500\text{ g}, ΔT=70∘C−20∘C=50∘C\Delta T = 70^\circ C - 20^\circ C = 50^\circ C, c=0.39 J g−1 ∘C−1c = 0.39 \text{ J g}^{-1} \text{ } ^\circ C^{-1}. Using Q=mcΔTQ = mc\Delta T: Q=500×0.39×50Q = 500 \times 0.39 \times 50 Q=25000×0.39=9750 JQ = 25000 \times 0.39 = 9750 \text{ J}.

Explanation:

Heat energy (QQ) is calculated by multiplying the mass, the specific heat capacity, and the temperature difference. Ensure that units for mass and specific heat are consistent (both in grams here).

Problem 3:

If the coefficient of cubical expansion (γ\gamma) of a metal is 4.8×10−5 ∘C−14.8 \times 10^{-5} \text{ } ^\circ C^{-1}, find its coefficient of linear expansion (α\alpha).

Solution:

We know the relationship: α=γ3\alpha = \frac{\gamma}{3}. Given γ=4.8×10−5 ∘C−1\gamma = 4.8 \times 10^{-5} \text{ } ^\circ C^{-1}. α=4.8×10−53=1.6×10−5 ∘C−1\alpha = \frac{4.8 \times 10^{-5}}{3} = 1.6 \times 10^{-5} \text{ } ^\circ C^{-1}.

Explanation:

For isotropic solids, the volume expansion coefficient is thrice the linear expansion coefficient. Dividing γ\gamma by 3 gives the linear expansion value.

Heat Transfer (Conduction, Convection, Radiation, Thermal Expansion) Revision - Class 8 Science ICSE