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The Particulate Nature of Matter - Thermal Energy Transfers

Grade 11IBPhysics

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

🔑Concepts

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Internal Energy (UU): The total energy of the molecules of a substance, which is the sum of the total random kinetic energy and the total intermolecular potential energy.

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Temperature (TT): A measure of the average random kinetic energy of the particles in a substance. In the absolute (Kelvin) scale, T∝Ek,avgT \propto E_{k,avg}.

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Thermal Equilibrium: Two objects are in thermal equilibrium when they are at the same temperature and there is no net heat transfer between them.

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Specific Heat Capacity (cc): The amount of energy required to raise the temperature of 1 kg1\ kg of a substance by 1 K1\ K (or 1∘C1^{\circ}C).

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Specific Latent Heat (LL): The energy required to change the phase of 1 kg1\ kg of a substance at constant temperature. LfL_f is for fusion (solid-liquid) and LvL_v is for vaporization (liquid-gas).

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Phase Changes: During a phase change, the temperature remains constant because the thermal energy supplied is used to increase the intermolecular potential energy (breaking bonds) rather than the kinetic energy.

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Conduction: The transfer of heat through solids via atomic vibrations and free electron collisions.

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Convection: The transfer of heat in fluids (liquids and gases) via the bulk movement of the fluid caused by density differences.

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Radiation: The transfer of energy via electromagnetic waves (infrared), which does not require a medium.

📐Formulae

Q=mcΔTQ = mc\Delta T

Q=mLQ = mL

T(K)=t(∘C)+273T(K) = t(^{\circ}C) + 273

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

C=QΔT=mcC = \frac{Q}{\Delta T} = mc

💡Examples

Problem 1:

An electric heater with a power rating of 1.5 kW1.5\ kW is used to heat 2.0 kg2.0\ kg of water from 20∘C20^{\circ}C to 80∘C80^{\circ}C. Calculate the time taken, assuming no heat loss to the surroundings. (Specific heat capacity of water c=4186 J kg−1K−1c = 4186\ J\ kg^{-1}K^{-1})

Solution:

Q=mcΔT=2.0×4186×(80−20)=2.0×4186×60=502,320 JQ = mc\Delta T = 2.0 \times 4186 \times (80 - 20) = 2.0 \times 4186 \times 60 = 502,320\ J t=QP=502,3201500≈335 st = \frac{Q}{P} = \frac{502,320}{1500} \approx 335\ s

Explanation:

First, calculate the total thermal energy required to change the temperature using Q=mcΔTQ = mc\Delta T. Then, use the power formula P=Q/tP = Q/t to solve for time, ensuring power is converted from kWkW to WW (1.5 kW=1500 W1.5\ kW = 1500\ W).

Problem 2:

Calculate the energy required to melt 0.5 kg0.5\ kg of ice at 0∘C0^{\circ}C and then heat the resulting water to 10∘C10^{\circ}C. (Specific latent heat of fusion of ice Lf=3.34×105 J kg−1L_f = 3.34 \times 10^5\ J\ kg^{-1}, specific heat of water c=4186 J kg−1K−1c = 4186\ J\ kg^{-1}K^{-1})

Solution:

Qtotal=Qmelt+QheatQ_{total} = Q_{melt} + Q_{heat} Qmelt=mLf=0.5×3.34×105=167,000 JQ_{melt} = mL_f = 0.5 \times 3.34 \times 10^5 = 167,000\ J Qheat=mcΔT=0.5×4186×(10−0)=20,930 JQ_{heat} = mc\Delta T = 0.5 \times 4186 \times (10 - 0) = 20,930\ J Qtotal=167,000+20,930=187,930 J≈1.88×105 JQ_{total} = 167,000 + 20,930 = 187,930\ J \approx 1.88 \times 10^5\ J

Explanation:

The process involves two stages: a phase change at constant temperature (melting) and a temperature increase. We sum the energy required for both stages using Q=mLQ = mL and Q=mcΔTQ = mc\Delta T respectively.

Thermal Energy Transfers - Revision Notes & Key Formulas | IB Grade 11 Physics