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Energy - Conduction, Convection, and Radiation

Grade 9IB

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

🔑Concepts

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Thermal Conduction: The process where thermal energy is transferred through a substance without any overall movement of the substance itself. It occurs primarily in solids via molecular vibrations and the movement of free electrons (in metals). The rate of conduction depends on the temperature gradient, ΔT\Delta T.

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Convection: The transfer of heat through fluids (liquids and gases) by the upward movement of the warmer, less dense regions of the fluid. As a fluid is heated, it expands, its density ρ\rho decreases, and it rises, creating a convection current.

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Thermal Radiation: The transfer of energy by infrared electromagnetic waves. Unlike conduction and convection, radiation does not require a medium and can occur in a vacuum. Surfaces that are dull and black are the best absorbers and emitters of radiation, while shiny, silver surfaces are the best reflectors.

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Specific Heat Capacity (cc): The amount of energy required to raise the temperature of 1 kg1\text{ kg} of a substance by 1∘C1^{\circ}\text{C} or 1 K1\text{ K}. It is a measure of how much thermal energy a material can store.

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Thermal Equilibrium: When two objects in thermal contact reach the same temperature, there is no net transfer of thermal energy between them.

📐Formulae

Q=mcΔTQ = mc\Delta T

ΔT=Tfinal−Tinitial\Delta T = T_{final} - T_{initial}

P=QtP = \frac{Q}{t}

ρ=mV\rho = \frac{m}{V}

💡Examples

Problem 1:

A 2.0 kg2.0\text{ kg} copper block is heated from 25∘C25^{\circ}\text{C} to 75∘C75^{\circ}\text{C}. If the specific heat capacity of copper is 390 J/kg∘C390\text{ J/kg}^{\circ}\text{C}, calculate the thermal energy absorbed by the block.

Solution:

Q=mcΔTQ = mc\Delta T ΔT=75∘C−25∘C=50∘C\Delta T = 75^{\circ}\text{C} - 25^{\circ}\text{C} = 50^{\circ}\text{C} Q=2.0×390×50Q = 2.0 \times 390 \times 50 Q=39000 JQ = 39000\text{ J}

Explanation:

The energy absorbed is calculated using the specific heat capacity formula. We first find the change in temperature ΔT\Delta T, then multiply the mass, specific heat capacity, and temperature change.

Problem 2:

An electric heater provides 1500 J1500\text{ J} of energy every second to a metal rod. If the rod reaches a steady state where it loses energy to the surroundings as fast as it gains it, what is the power of the heat loss?

Solution:

Pin=PoutP_{in} = P_{out} Pout=1500 WP_{out} = 1500\text{ W}

Explanation:

At a steady state (thermal equilibrium with the environment's heating rate), the rate of energy gain equals the rate of energy loss. Since 1 J/s=1 W1\text{ J/s} = 1\text{ W}, the power loss via conduction, convection, and radiation is 1500 W1500\text{ W}.

Problem 3:

A student compares two flasks filled with hot water at 90∘C90^{\circ}\text{C}. Flask A is painted matte black and Flask B is painted shiny silver. Which flask will cool down faster and why?

Solution:

Flask A (matte black) will cool down faster.

Explanation:

Flask A is a better emitter of infrared radiation because matte black surfaces have higher emissivity than shiny silver surfaces. Therefore, Flask A loses thermal energy to the surroundings more rapidly through radiation.