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Thermal Properties of Matter - Heat Transfer (Conduction, Convection, and Radiation)

Grade 11CBSEPhysics

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

πŸ”‘Concepts

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Heat transfer is the flow of thermal energy from a body at a higher temperature to a body at a lower temperature via three modes: Conduction, Convection, and Radiation.

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Conduction is the process of heat transfer in solids through molecular vibrations without the actual movement of the particles. The rate of heat flow depends on the material's thermal conductivity kk.

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Convection involves the transfer of heat by the actual movement of the fluid (liquid or gas) particles. It can be natural (buoyancy-driven) or forced (pump/fan-driven).

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Radiation is the transfer of energy through electromagnetic waves and does not require a material medium. All bodies above 0Β K0\text{ K} emit thermal radiation.

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Thermal Resistance RR is analogous to electrical resistance and is defined as the ratio of temperature difference to the rate of heat flow: R=Ξ”THR = \frac{\Delta T}{H}.

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A Black Body is an ideal surface that absorbs all incident electromagnetic radiation. Its emissive power is governed by the Stefan-Boltzmann Law.

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Wien's Displacement Law states that the wavelength Ξ»m\lambda_m corresponding to maximum spectral emissive power is inversely proportional to the absolute temperature TT.

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Newton's Law of Cooling states that the rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings, provided the difference is small.

πŸ“Formulae

H=dQdt=kAT1βˆ’T2LH = \frac{dQ}{dt} = kA \frac{T_1 - T_2}{L}

R=LkAR = \frac{L}{kA}

E=ΟƒeAT4E = \sigma e A T^4

Ξ»mT=b\lambda_m T = b

dTdt=βˆ’K(Tβˆ’Ts)\frac{dT}{dt} = -K(T - T_s)

T1βˆ’T2t=K[T1+T22βˆ’Ts]\frac{T_1 - T_2}{t} = K \left[ \frac{T_1 + T_2}{2} - T_s \right]

πŸ’‘Examples

Problem 1:

A metal rod of length 0.5Β m0.5\text{ m} and cross-sectional area 0.02Β m20.02\text{ m}^2 has its ends maintained at 100∘C100^{\circ}\text{C} and 0∘C0^{\circ}\text{C}. If the thermal conductivity kk is 400Β WΒ mβˆ’1Kβˆ’1400\text{ W m}^{-1}\text{K}^{-1}, find the rate of heat flow.

Solution:

Given: L=0.5Β mL = 0.5\text{ m}, A=0.02Β m2A = 0.02\text{ m}^2, T1=100∘CT_1 = 100^{\circ}\text{C}, T2=0∘CT_2 = 0^{\circ}\text{C}, k=400Β WΒ mβˆ’1Kβˆ’1k = 400\text{ W m}^{-1}\text{K}^{-1}. Using the formula: H=kA(T1βˆ’T2)LH = \frac{kA(T_1 - T_2)}{L} H=400Γ—0.02Γ—(100βˆ’0)0.5H = \frac{400 \times 0.02 \times (100 - 0)}{0.5} H=8Γ—1000.5=1600Β WH = \frac{8 \times 100}{0.5} = 1600\text{ W}

Explanation:

The rate of heat flow HH is calculated using the steady-state conduction formula, where heat flows from the higher temperature end to the lower temperature end.

Problem 2:

A body cools from 60∘C60^{\circ}\text{C} to 50∘C50^{\circ}\text{C} in 1010 minutes. If the surrounding temperature is 25∘C25^{\circ}\text{C}, find the time it will take to cool from 50∘C50^{\circ}\text{C} to 40∘C40^{\circ}\text{C}.

Solution:

Case 1: T1=60,T2=50,t=10,Ts=25T_1 = 60, T_2 = 50, t = 10, T_s = 25. 60βˆ’5010=K(60+502βˆ’25)β€…β€ŠβŸΉβ€…β€Š1=K(55βˆ’25)β€…β€ŠβŸΉβ€…β€ŠK=130\frac{60 - 50}{10} = K \left( \frac{60 + 50}{2} - 25 \right) \implies 1 = K(55 - 25) \implies K = \frac{1}{30}. Case 2: T1=50,T2=40,Ts=25,t=?T_1 = 50, T_2 = 40, T_s = 25, t = ?. 50βˆ’40t=130(50+402βˆ’25)\frac{50 - 40}{t} = \frac{1}{30} \left( \frac{50 + 40}{2} - 25 \right) 10t=130(45βˆ’25)β€…β€ŠβŸΉβ€…β€Š10t=2030β€…β€ŠβŸΉβ€…β€Št=15Β minutes\frac{10}{t} = \frac{1}{30} (45 - 25) \implies \frac{10}{t} = \frac{20}{30} \implies t = 15\text{ minutes}.

Explanation:

We apply the average form of Newton's Law of Cooling for two different intervals to find the constant KK and then the unknown time tt.

Heat Transfer (Conduction, Convection, and Radiation) Revision - Class 11 Physics CBSE