UNIT 2

CLASS 11 PHYSICS • HEAT AND THERMODYNAMICS

Rate of Heat Flow

Chapter 12

Original Scanned PDF – View Notes

Thermal Conductivity

Consider a solid block of thickness x and cross-sectional area A. Let its two opposite faces be maintained at temperatures T₁ and T₂, where T₁ > T₂. Heat flows from the hotter face toward the colder face.

Heat flows from the face at T₁ to the face at T₂.

Experimentally, the amount of heat Q transferred in time t is found to be:

i. Directly proportional to the area:

Q ∝ A … (i)

ii. Directly proportional to the temperature difference:

Q ∝ (T₁ − T₂) … (ii)

iii. Directly proportional to the time:

Q ∝ t … (iii)

iv. Inversely proportional to the distance between the faces:

Q ∝ 1/x … (iv)

Combining all four relations:

Q ∝ A(T₁ − T₂)t/x

Q = KA(T₁ − T₂)t/x

Therefore:

Q/t = KA(T₁ − T₂)/x

Thermal conductivity, K: The proportionality constant in the heat-conduction equation. Its value depends on the nature of the material.

Numerically, the thermal conductivity of a material is equal to the rate of heat flow between two faces of area 1 m², separated by 1 m, when the temperature difference between them is 1°C or 1 K.

SI unit: W m⁻¹ K⁻¹.

Modes of Transfer of Heat

Three modes of heat transfer represented in the source notes.

1. Conduction

The mode of transfer of heat in which heat is transferred between two points without actual movement of particles is called conduction.

In this mode, heat spreads to neighbouring particles and is transferred successively from particle to particle. The source gives heat transfer in solids as an example.

2. Convection

The mode of transfer of heat in which heat is transferred between two points by actual movement of particles is called convection.

The source gives heat transfer in liquids and gases as examples.

3. Radiation

The mode of transfer of heat without the presence of any material medium, in the form of infrared waves, is called radiation.

Example given in the scan: transfer of heat from the Sun to the Earth’s surface.

Absorption, Transmission and Reflection Coefficients

When heat radiation is incident on a body, it may be partly absorbed, partly transmitted and partly reflected.

Incident radiation Q is divided into absorbed A, transmitted T and reflected R components.

Let Q be the amount of incident heat radiation and A, T and R be the absorbed, transmitted and reflected amounts respectively.

Q = A + T + R

Absorption Coefficient, a

The ratio of heat radiation absorbed to heat radiation incident is called the absorption coefficient.

a = A/Q ⇒ A = aQ

Transmission Coefficient, t

The ratio of heat radiation transmitted to heat radiation incident is called the transmission coefficient.

t = T/Q ⇒ T = tQ

Reflection Coefficient, r

The ratio of heat radiation reflected to heat radiation incident is called the reflection coefficient.

r = R/Q ⇒ R = rQ

Using Q = A + T + R:

Q = aQ + tQ + rQ

a + t + r = 1

Black Body

A body which completely absorbs heat radiation of all wavelengths falling on it is called a perfectly black body.

For a perfectly black body, the absorption coefficient is unity. The source notes mention the Sun as approximately behaving like a black body even though it appears bright, and state that a black body absorbs about 96% to 98% of incident radiation.

Fery’s Black Body

Fery’s black body is an experimental model consisting of a closed, double-walled hollow sphere with a small hole O and a conical projection P opposite the hole.

Fery’s black body: repeated internal reflection leads to almost complete absorption.

Radiation entering through the small hole undergoes repeated reflection from the inner walls and the conical projection. Almost all the radiation is eventually absorbed. When the black body is heated, radiation emerges from hole O.

Emissive Power

The emissive power of a body is the amount of heat radiation emitted per second per unit area by the body over all wavelengths.

E = Q/(At)

SI unit: W m⁻².

Emissivity, e

The ratio of the emissive power of a body to the emissive power of a perfectly black body at the same temperature is called emissivity.

e = Emissive power of body / Emissive power of perfectly black body

i. For e < 1, the body is not a perfectly black body.

ii. For e = 1, the body is a perfectly black body.

iii. For e > 1, the condition is impossible.

Stefan–Boltzmann Law

The heat radiation emitted per second per unit area by a perfectly black body is directly proportional to the fourth power of its absolute temperature.

E ∝ T⁴

E = σT⁴

Here σ is the Stefan–Boltzmann constant.

σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴

For a body of emissivity e:

E = eσT⁴

If a body at absolute temperature T₁ is surrounded by another body or enclosure at temperature T₂, the net emissive power is:

E = eσ(T₁⁴ − T₂⁴)

Measurement of Thermal Conductivity of a Solid by Searle’s Method

Searle’s apparatus, corresponding to pages 6–7 of the scanned notes.

The apparatus consists of a uniform experimental rod of cross-sectional area A with two cavities C₁ and C₂ separated by distance x. One end is heated by steam. Water is continuously circulated through a pipe around the other end. Thermometers T₁ and T₂ measure the temperatures at the two cavities, while T₃ and T₄ measure the outlet and inlet temperatures of water respectively.

The rod is surrounded by insulating material such as wood or cotton to reduce heat exchange with the surroundings. After steady state is reached, all thermometer readings become constant.

Rate of heat conducted between C₁ and C₂:

Q/t = KA(T₁ − T₂)/x … (i)

If m is the mass of water collected per second and s is the specific heat capacity of water, heat absorbed per second by water is:

Q/t = ms(T₃ − T₄) … (ii)

At steady state, equations (i) and (ii) are equal:

KA(T₁ − T₂)/x = ms(T₃ − T₄)

K = ms(T₃ − T₄)x / [A(T₁ − T₂)]

Conduction Numericals from the Scanned Notes

Numerical 1 – Ice Melted by Heat Flow through a Bar

Question: A bar 0.2 m long and 2.5 cm² in cross-section is ideally lagged. One end is maintained at 100°C and the other at 0°C by immersing it in melting ice. Calculate the mass of ice melted in one hour. Thermal conductivity of the bar is 4 × 10⁻² W m⁻¹ K⁻¹.

x = 0.2 m, A = 2.5 × 10⁻⁴ m², T₁ = 100°C, T₂ = 0°C, t = 3600 s

K = 4 × 10⁻² W m⁻¹ K⁻¹

Heat conducted = heat used to melt ice:

mL_f = KA(T₁ − T₂)t/x

m = 5.36 × 10⁻⁵ kg

Numerical 2 – Aluminium and Brass Composite Rod

Question: A rod 1.3 m long consists of 0.8 m of aluminium joined end-to-end to 0.5 m of brass. The aluminium end is at 150°C and the brass end at 20°C. No heat is lost from the sides. Find the steady-state temperature at the junction.

x_Al = 0.8 m, x_Br = 0.5 m

K_Al = 205 W m⁻¹ K⁻¹, K_Br = 110 W m⁻¹ K⁻¹

At steady state, rates of heat flow are equal:

205(150 − T)/0.8 = 110(T − 20)/0.5

256.25(150 − T) = 220(T − 20)

T = 89.94°C ≈ 90°C

Numerical 3 – Thermal Conductivity of a Stone Slab

Question: A stone slab of area 0.36 m² and thickness 10 cm is exposed on its lower surface to steam at 100°C. A block of ice at 0°C rests on the upper surface. In one hour, 4.8 kg of ice is melted. Calculate the thermal conductivity of the stone.

A = 0.36 m², x = 0.1 m, t = 3600 s, m = 4.8 kg

mL_f/t = K A(100 − 0)/x

K = 1.244 W m⁻¹ K⁻¹

Numerical 4 – Thermal Conductivity of a Metal Rod

Question: A metal rod 20 cm long and of cross-sectional area 3.14 cm² is covered with a non-conducting material. One end is maintained at 100°C while the other end is kept in ice at 0°C. If 25 g of ice melts in 5 minutes, calculate the thermal conductivity of the metal.

x = 0.2 m, A = 3.14 × 10⁻⁴ m²

m = 25 × 10⁻³ kg, t = 300 s

mL_f/t = KA(100 − 0)/x

K = 178.34 W m⁻¹ K⁻¹

Numerical 5 – Heat Loss through a Glass Window

Question: Estimate the rate of heat loss through a glass window of area 2 m² and thickness 4 mm when the room temperature is 300 K and the outside temperature is 5°C. The source uses K = 1.2 W m⁻¹ K⁻¹.

A = 2 m², x = 4 × 10⁻³ m

T₁ = 300 K = 27°C, T₂ = 5°C

Q/t = KA(T₁ − T₂)/x

Q/t = 13200 W

Numerical 6 – Interface Temperature in a Wood–Cork Ice Box

Question: An ice box is made of wood 1.75 cm thick and lined inside with cork 2 cm thick. The temperature of the inner surface is steady and the thermal conductivity of wood is five times that of cork. Find the interface temperature using the source conditions.

x_wood = 0.0175 m, x_cork = 0.02 m, K_wood = 5K_cork

The source sets the steady-state heat-flow rates equal and obtains:

0.24 = 0.0875(12 − T)

T = 9.25°C

Numerical 7 – Heat Loss from a Hand through a Woollen Glove

Question: Assuming that the insulation provided by a woollen glove is equivalent to a layer of quiescent air 3 mm thick, determine the heat loss per minute from a man’s hand of surface area 200 cm² on a winter day when the air temperature is −3°C. The skin temperature is 35°C and thermal conductivity of air is 24 × 10⁻³ W m⁻¹ K⁻¹.

x = 3 × 10⁻³ m, A = 0.02 m²

ΔT = 35 − (−3) = 38°C

Q/t = (24 × 10⁻³)(0.02)(38)/(3 × 10⁻³)

Q/t = 6.08 J s⁻¹

Heat loss per minute = 364.8 J min⁻¹

Numerical 8 – Glass Window, 3 mm Thick

Question: Estimate the rate of heat loss through a glass window of area 2 m² and thickness 3 mm when the room temperature is 20°C and outside temperature is 5°C. Given K = 1.2 W m⁻¹ K⁻¹.

Q/t = 1.2 × 2 × (20 − 5)/(3 × 10⁻³)

Q/t = 12000 W = 12 kW

Numerical 9 – Rate of Melting Ice in a Wooden Box

Question: Estimate the rate at which ice melts in a wooden box 2.5 cm thick with inside dimensions 100 cm × 60 cm × 40 cm. The external temperature is 35°C and thermal conductivity of wood is 0.168 W m⁻¹ K⁻¹.

x = 2.5 × 10⁻² m

A = 2(lb + bh + hl) = 2.48 m²

K = 0.168 W m⁻¹ K⁻¹

m/t = KA(35)/(xL_f)

m/t = 1.736 × 10⁻³ kg s⁻¹

Numerical 10 – Rate at Which Ice Melts at the End of a Bar

Question: A bar 0.2 m long with cross-sectional area 2.5 × 10⁻⁴ m² is ideally lagged. One end is at 373 K and the other at 273 K in melting ice. Calculate the rate at which ice melts, using K = 4 × 10⁻² W m⁻¹ K⁻¹.

m/t = KA(T₁ − T₂)/(xL_f)

m/t = 1.48 × 10⁻⁸ kg s⁻¹

Numerical 11 – Temperature of the Lower Surface of a Steel Pot

Question: A pot with a steel bottom 8.5 mm thick rests on a hot surface. The bottom area is 0.15 m². Water in the pot is at 100°C and 390 g evaporates every 3 minutes. Find the temperature of the lower surface of the pot. The source uses K = 50.2 W m⁻¹ K⁻¹ and L_v = 2256 × 10³ J kg⁻¹.

x = 0.0085 m, A = 0.15 m², m = 0.39 kg, t = 180 s

KA(T₁ − 373)/x = mL_v/t

T₁ = 378.52 K = 105.52°C

Radiation Numericals from the Scanned Notes

Numerical 12 – Temperature of an Electric Fire Element

Question: The element of an electric fire with output 1.5 kW is a cylinder 0.3 m long and 0.04 m in radius. Calculate its temperature if it behaves as a black body.

P = 1500 W, l = 0.3 m, r = 0.04 m

Curved surface area used in the source: A = 2πrl ≈ 7.5 × 10⁻² m²

P = AσT⁴

T ≈ 769.64 K

Numerical 13 – Power Radiated by the Sun and Solar Flux at Earth

Question: The Sun is treated as a black body of surface temperature about 6000 K. If its radius is 7 × 10⁸ m, calculate the energy per second radiated from its surface. The Earth is about 1.5 × 10¹¹ m from the Sun. Assuming the radiation spreads over a sphere of that radius, estimate the energy received per second per square metre at Earth.

Surface area of Sun = 4πR²

P = σAT⁴

P ≈ 4.52 × 10²⁶ W

At Earth’s orbit, area = 4πd²

Solar flux = P/(4πd²)

Solar flux ≈ 1598.62 W m⁻²

Numerical 14 – Experimental Value of Stefan’s Constant

Question: A sphere of radius 2.00 cm with a black surface is cooled and suspended in a large evacuated enclosure with black walls maintained at 27°C. If the rate of change of thermal energy is 1.85 J s⁻¹ when the sphere is at −73°C, calculate Stefan’s constant.

r = 0.02 m, T₁ = 300 K, T₂ = 200 K

P = σA(T₁⁴ − T₂⁴)

σ = 1.85 / {4π(0.02)²[(300)⁴ − (200)⁴]}

σ ≈ 5.66 × 10⁻⁸ W m⁻² K⁻⁴

Numerical 15 – Power Loss from Unit Area of a Black Body

Question: Estimate the power loss through unit area from a perfectly black body at 327°C to surroundings at 27°C.

A = 1 m², T₁ = 600 K, T₂ = 300 K

P = σA(T₁⁴ − T₂⁴)

P = 6889.05 W

Numerical 16 – Radiant Power of a Spherical Black Body

Question: A spherical black body of radius 5 cm has temperature 127°C and emissivity 0.6. Calculate its radiant power.

r = 0.05 m, A = 4πr² ≈ 0.0314 m²

T = 400 K, e = 0.6

P = eσAT⁴

P = 27.36 W

Numerical 17 – Radiant Power Loss from a Human Body

Question: Estimate the radiant power loss from a human body at 38.5°C to an environment at 0°C if the body surface area is 1.5 m² and emissivity is 0.6.

T₁ = 311.5 K, T₂ = 273 K, A = 1.5 m², e = 0.6

P = eσA(T₁⁴ − T₂⁴)

P = 197.01 W

Numerical 18 – Rate of Heat Loss from a Man’s Body

Question: A man whose surface area is 2 m² is sitting in a room where the air temperature is 20°C. If his skin temperature is 37°C and the emissivity of his skin is 0.97, find the rate at which his body loses heat.

A = 2 m², T_skin = 310 K, T_room = 293 K, e = 0.97

P = Aeσ(T_skin⁴ − T_room⁴)

P = 205.16 W

Numerical 19 – Radiation Ratio at 2500 K and 2000 K

Question: What is the ratio of the energy per second radiated by the filament of a lamp at 2500 K to that radiated at 2000 K, assuming the filament is a black-body radiator?

P ∝ T⁴

P₁/P₂ = (2500/2000)⁴

P₁ : P₂ = 2.44 : 1