NEET Physics Class 11 Chapter 8 Slabs In Parallel And Series

NEET Physics Class 11 Chapter 8 Slabs In Parallel And Series

Slabs In Series (in steady state)

Consider a composite slab consisting of two materials having different thicknesses L₁ and L₂, different cross-sectional areas A₁ and A_2, and different thermal conductivities K₁ and K₂. The temperature at the outer surface at the ends is maintained at T_H and T_C, and all lateral surfaces are covered by an adiabatic coating.

Let the temperature at the junction be T, since a steady state has been achieved thermal current through each slab will be equal. Then thermal current through the first slab.

⇒ \(\mathrm{i}=\frac{Q}{t}=\frac{T_H-T}{R_1}\)

or \(\mathrm{T}_{\mathrm{H}}-\mathrm{T}=\mathrm{iR} \mathrm{R}_1\)……(1)

and that through the second slab,

⇒ \(\mathrm{i}=\frac{Q}{t}\)

⇒ \(\frac{T-T_C}{R_2}\)

or \(\mathrm{T}-\mathrm{T}_{\mathrm{c}}=\mathrm{i} \mathrm{R}_2\)…..(2)

adding eqn. (1) and eqn (2)

T_H– T_L= (R₁+ R₂) i

or \(\mathrm{i}=\frac{T_H-T_C}{R_1+R_2}\)

Thus these two slabs are equivalent to a single slab of thermal resistance R1+ R2. If more than two slabs are joined in series and are allowed to attain a steady state, then equivalent thermal resistance is given by

R = R₁+ R₂+ R₃+ ……. (3)

Question 1. The figure shows the cross-section of the outer wall of a house built in a hill resort to keep the house insulated from the freezing temperature outside. The wall consists of teak wood of thickness L₁ and brick of thickness (L₂= 5L₁), sandwiching two layers of an unknown material with identical thermal conductivities and thickness. The thermal conductivity of teak wood is K1and that of brick is (K₂= 5K). Heat conduction through the wall has reached a steady state with the temperature of three surfaces being known. (T₁= 25ºC, T₂= 20ºC and T₅= –20ºC). Find the interface temperature T₄ and T₃.

Answer: Let the interface area be A. Then thermal resistance of wood,

⇒ \(\mathrm{R}_1=\frac{L_1}{K_1 A}\)

and that of brick wall \(\mathrm{R}_2=\frac{L_2}{K_2 A}=\frac{5 L_1}{5 K_1 A}=\mathrm{R}_1\)

Let the thermal resistance of each sandwiched layer = R. Then the above wall can be visualized as a circuit

thermal current through each wall is the same.

Hence \(\frac{25-20}{R_1}=\frac{20-T_3}{R}\)

⇒ \(\frac{T_3-T_4}{R}=\frac{T_4+20}{R_1}\)

⇒ 25 – 20 = T4+ 20

⇒ T₄= –15ºC

also, 20 – T₃= T₃– T₄

⇒ \(\mathrm{T}_3=\frac{20+T_4}{2}\)

= 2.5ºC

Question 2. In example 3, K₁= 0.125 W/m–ºC, K₂= 5K₁= 0.625 W/m–ºC, and the thermal conductivity of the unknown material is K = 0.25 W/mºC. L₁= 4cm, L₂= 5L₁= 20cm and L = 10cm. If the house consists of a single room with a total wall area of 100 m₂, then find the power of the electric heater being used in the room.
Answer:

R₁ = R₂ = \(\frac{\left(4 \times 10^{-2} \mathrm{~m}\right)}{\left(0.125 \mathrm{w} / \mathrm{m}-{ }^{\circ} \mathrm{C}\right)\left(100 \mathrm{~m}^2\right)}\)

= 32 × 10^-4ºC/w

R = \(\frac{\left(10 \times 10^{-2} \mathrm{~m}\right)}{\left(0.25 \mathrm{~W} / \mathrm{m}-{ }^{\circ} \mathrm{C}\left(100 \mathrm{~m}^2\right)\right.}\)

= 40 × 10^-4ºC/w

The equivalent thermal resistance of the entire wall = R₁+ R₂+ 2R = 144×10^-4ºC/W

∴ Net heat current, i.e. amount of heat flowing out of the house per second

⇒ \(\frac{T_H-T_C}{R}=\frac{25^{\circ} \mathrm{C}-\left(-20^{\circ} \mathrm{C}\right)}{144 \times 10^{-40} \mathrm{C} / \mathrm{w}}=\frac{45 \times 10^4}{144}\)watt = 3.12 Kwatt

Hence the heater must supply 3.12 kW to compensate for the outflow of heat.

Slabs In Parallel :

Consider two slabs held between the same heat reservoirs, their thermal conductivities K₁ and K_2, and cross-sectional areas A₁ and A₂ Heat reservoir

then \(\mathrm{R}_1=\frac{L}{K_1 A_1}\), \(\mathrm{R}_2=\frac{L}{K_2 A_2}\)

thermal current through slab 1 and that through slab 2 Net heat current from the hot to cold reservoir

⇒ \(i_1=\frac{T_H-T_C}{R_1}\)

Comparing with i = \(i_1+i_2=\left(T_H-T_C\right)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\),

⇒ \(\mathrm{i}=\frac{T_H-T_C}{R_{e q}}\) we get, \(\)

⇒ \(\frac{1}{R_{c q}}=\frac{1}{R_1}+\frac{1}{R_2}\)

If more than two rods are joined in parallel, the equivalent thermal resistance is given by

⇒ \(\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)

Question 3. Figure shows a copper rod joined to a steel rod. The rods have equal length and equal cross sectional area. The free end of the copper rod is kept at 0ºC and that of the steel rod is kept at 100ºC. Find the temperature of the junction of the rod. Conductivity of copper = 390 W/mºC Conductivity of steel = 46 W/m ºC 0ºC Copper Steel 100ºC
Answer:

Heat current in first rod (copper) = \(\frac{390 \times A(A-\theta)}{\ell}\)

Here θ is the temperature of the junction and A and l are the area and length of the copper rod.

Heat current in second rod (steel) = \(\frac{46 \times A(\theta-100)}{\ell}\)

In series combination. heat current remains the same. So,

⇒ \(\frac{390 \times A(0-\theta)}{\ell}\)

⇒ \(\frac{46 \times A(\theta-100)}{\ell}\)

⇒ -390 θ = 46θ-4600

⇒ 436 θ = 4600 θ = 10.6ºC

Question 4. An aluminum rod and a copper rod of equal length 1m and cross-sectional area 1cm₂ are welded together as shown in the figure. One end is kept at a temperature of 20ºC and the other at 60ºC. Calculate the amount of heat taken out per second from the hot end. The thermal conductivity of aluminum is 200 W/mºC and of copper is 390 W/mºC.

⇒ \(20^{\circ} \mathrm{C} \begin{array}{|l|}
\hline \text { Aluminium } \\
\hline \text { Copper } \\
\hline\end{array} 60^{\circ} \mathrm{C}\)

Answer: Heat current through the \(\frac{200 \times\left(1 \times 10^{-4}\right)}{1}\) = (60-20)

Heat current through the copper rod = \(\frac{390 \times\left(1 \times 10^{-4}\right)}{1}\) . (60-20)

Total heat = 200 × 10–4 × 40 + 390 × 10–4× 40 = 590 × 40 × 10–4= 2.36 Joule

Question 5. The three rods shown in the figure have identical geometrical dimensions. Heat flows from the hot end at the rate of 40W in an arrangement

1. Find the rate of heat flow when the rods are joined in arrangement
2. The thermal conductivity of aluminum and copper are 200 W/mºC and 400 W/mºC respectively.

Answer:

In the arrangement

The three rods are joined in series. The rate of flow of heat,

⇒ \(\frac{d \theta}{d t}=\frac{K A\left(\theta_1-\theta_2\right)}{\ell}=\frac{\theta_1-\theta_2}{R}\)

But, R = R₁+ R₂+ R₃[In series]

∴ 40 = \(\frac{100-0}{R_1+R_2+R_1}\)

40 = \(\frac{100}{\frac{\ell}{K_1 A}+\frac{\ell}{K_2 A}+\frac{\ell}{K A}}\)

40 = \(\frac{100}{\frac{\ell}{A}\left[\frac{2}{K_1}+\frac{1}{K_2}\right]}\)

⇒ \(\frac{\ell}{A}\left[\frac{2}{200}+\frac{1}{400}\right]=\frac{100}{40}\)

⇒ \(\frac{\ell}{A}\)

= 200 per m

In Figure two rods are all in parallel and the resultant of both is in series with the first rod

∴ \(\frac{d Q}{d t}=\frac{\theta_1-\theta_2}{R}\)

But R = \(\mathrm{R}_1+\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}\)

⇒ \(\frac{d Q}{d t}=\frac{100-0}{\frac{\ell}{K_1 A}+\frac{1}{\frac{K_1 A}{\ell}+\frac{K_2 A}{\ell}}}\)

⇒ \(\frac{100-0}{\frac{\ell}{A}\left[\frac{1}{K_1}+\frac{1}{K_1+K_2}\right]}\)

⇒ \(\frac{600 \times 100}{200 \times 4}\)

= 75W

Question 6. A metal rod of length 20cm and diameter 2 cm is covered with a nonconducting substance. One of its ends is maintained at 100ºC while the other end is put at 0ºC. It is found that 25 g of ice melts in 5 min. Calculate the coefficient of thermal conductivity of the metal. Latent heat of ice = 80 cal gram^-1
Answer:

Here, the length of the rod,

Δx = 20 cm = 20 × 10^-2m

Diameter = 2cm,

Radius = r = 1 cm = 10^-2m

Area of cross-section

a = πr²= π(10^-2)²= π × 10^-4 sq. m

ΔT = 100 – 0 = 100ºC

Mass of ice melted, m = 25g

As L = 80 cal/ g

∴ Heat conducted, ΔQ = mL = 25 × 80 = 2000 cal = 2000 × 4.2 J

Δt = 5 min = 300 s

From = \(\frac{\Delta Q}{\Delta t}=\mathrm{KA} \frac{\Delta T}{\Delta x}\)

K = \(\frac{2000 \times 4.2 \times 20 \times 10^{-2}}{300 \times 10^{-4} \pi \times 100}\)

= 1.78Js^-1m^-1ºC^-1

Question 7. Two thin concentric shells made from copper with radius r₁ and r₂(r₂> r₁) have a material of thermal conductivity K filled between them. The inner and outer spheres are maintained at temperatures T_H and T_C respectively by keeping a heater of power P at the center of the two spheres. Find the value of P.
Answer:

Heat flowing per second through each cross-section of the sphere = P = i. Thermal resistance of the spherical shell of radius x and thickness dx,

dR = \(\frac{d x}{K .4 \pi x^2}\)

⇒ \(\mathrm{R}=\int_n^2 \frac{d x}{4 \pi x^2 \cdot K}=\frac{1}{4 \pi K}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

thermal current i = P = \(\frac{T_H-T_C}{R}=\frac{4 \pi K\left(T_H-T_C\right) r_1 r_2}{\left(r_2-r_1\right)}\)