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Thermal Resistance | Thermal Resistance Formula

Thermal resistance is a concept that quantifies how effectively a material or an object resists the flow of heat. It represents the impediment that a substance poses to the transfer of heat energy. Just as electrical resistance hinders the flow of electric current, thermal resistance hampers the flow of heat. This concept is fundamental in understanding and designing heat transfer systems and insulation.

Thermal Resistance Formula

In the article Fourier Law of Heat Conduction, the heat current(dQ/dt) is given by

$\displaystyle \frac{Q}{t}=\frac{kA({{\theta }_{1}}-{{\theta }_{2}})}{l}$

$\displaystyle \Rightarrow \frac{Q}{t}=\frac{({{\theta }_{1}}-{{\theta }_{2}})}{\left( {}^{l}\!\!\diagup\!\!{}_{Ak}\; \right)}$ …..(1)

Above equation is mathematically equivalent to OHM’s law, with temperature difference playing the role of electric potential difference(V) and heat current(dQ/dt) playing the role of electric current(I). So comparing equation(1) with OHM’s law, $\displaystyle I=\frac{V}{R}$ , we get

$\displaystyle \mathbf{Thermal}\text{ }\mathbf{Resistance}(R)=\frac{l}{Ak}$ …..(2)

Hence for a slab of cross-section A, lateral thickness l and thermal conductivity k, thermal resistance is given by equation(2).

Unit of thermal resistance

S.I. unit of thermal resistance is kelvins per watt (K/W) or degrees Celsius per watt (^oC/W)

Note :-

Kirchoff’s current law: We have seen earlier that, for a slab in steady state, the thermal current(dQ/dt) remains same at each cross-section. This is analogous to Kirchoff’s current law in electricity, which can now be very conveniently applied to thermal conduction.
Composite Systems and Total Resistance: For complex systems involving multiple layers or materials, the total thermal resistance is the sum of the individual resistances. This is analogous to adding resistors in an electrical circuit. For example, in a wall with layers of insulation and other materials, the total thermal resistance can be calculated by summing up the thermal resistances of each layer.

Example 1.

Three identical rods of length 1 m each, having cross-section area of 1 cm² each and made of aluminium, copper and steel respectively are maintained at temperatures of 12 ºC, 4 ºC and 50 ºC respectively at their separate ends. Find the temperature of their common junction.
[ k_Cu = 400 W/m-K , k_Al = 200 W/m-K , k_steel = 50 W/m-K ]

Solution:

$\displaystyle {{R}_{Al}}=\frac{l}{Ak}=\frac{1}{{{10}^{-4}}\times 200}=\frac{{{10}^{4}}}{200}$

Similarly,

$\displaystyle {{R}_{Steel}}=\frac{{{10}^{4}}}{50}$ and $\displaystyle {{R}_{Cu}}=\frac{{{10}^{4}}}{400}$

Let temperature of common junction = T, then from Kirchhoff’s current law,

$\displaystyle {{\left( \frac{Q}{t} \right)}_{Al}}+{{\left( \frac{Q}{t} \right)}_{Steel}}+{{\left( \frac{Q}{t} \right)}_{Cu}}=0$

Thermal Resistance | Thermal Resistance Formula

$\displaystyle \Rightarrow \frac{T-12}{{{R}_{Al}}}+\frac{T-50}{{{R}_{Steel}}}+\frac{T-4}{{{R}_{Cu}}}=0$

$\displaystyle \Rightarrow (T-12)\times 200+(T-50)\times 50+(T-4)\times 400=0$

$\displaystyle \Rightarrow 4(T-12)+(T-50)+8(T-4)=0$

$\displaystyle \Rightarrow 13T=48+50+32$

$\displaystyle \Rightarrow 13T=130$

$\displaystyle \Rightarrow T=10{}^{\circ }C$

Slabs in Series and Parallel

(Thermal Resistance)

When dealing with thermal resistance of composite systems, to find out equivalent thermal resistance, the concept of series and parallel combinations applies. Let’s discuss how these concepts work for slabs in series and parallel.

1. Series Thermal Resistance: In a series arrangement, multiple slabs are placed one after the other, creating a continuous path for heat to flow through. The total thermal resistance of the series combination is the sum of the individual thermal resistances of each slab.

Let us consider a composite slab consisting of two materials having different thicknesses l₁ and l₂, different cross-sectional areas A₁ and A₂ and different thermal conductivities k₁ and k₂. The temperature at the outer surface of the slabs are maintained at T_H and T_L, and all lateral surfaces are covered by an adiabatic layer.

Let temperature at the junction be T and steady state has been achieved. In that case thermal current through each slab will be equal.

Thermal current through the first slab :

$\displaystyle \frac{Q}{t}=\frac{({{T}_{H}}-T)}{{{R}_{1}}}$

$\displaystyle \Rightarrow {{T}_{H}}-T=\left( \frac{Q}{t} \right){{R}_{1}}$ …..(3)

Similarly thermal current through the second slab :

$\displaystyle \frac{Q}{t}=\frac{(T-{{T}_{L}})}{{{R}_{2}}}$

$\displaystyle \Rightarrow T-{{T}_{L}}=\left( \frac{Q}{t} \right){{R}_{2}}$ …..(4)

Adding eq. (3) and (4), we get,

$\displaystyle {{T}_{H}}-{{T}_{L}}=\left( {{R}_{1}}+{{R}_{2}} \right)\frac{Q}{t}$

$\displaystyle \Rightarrow \frac{Q}{t}=\frac{{{T}_{H}}-{{T}_{L}}}{{{R}_{1}}+{{R}_{2}}}$ …..(5)

From eq. (5) we conclude that 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 steady state, then equivalent thermal resistance is given by :-

$\displaystyle {{R}_{Series}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+.....$ …..(6)

Temperature of Junction Point

Let T bet the temperature of junction point. Now since at steady state, the thermal current through both the slabs is equal, hence

$\displaystyle \frac{Q}{t}=\frac{({{T}_{H}}-T)}{{{R}_{1}}}=\frac{(T-{{T}_{L}})}{{{R}_{2}}}$

$\displaystyle \Rightarrow \frac{{{T}_{H}}}{{{R}_{1}}}+\frac{{{T}_{L}}}{{{R}_{2}}}=T\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)$

$\displaystyle \Rightarrow \frac{{{T}_{H}}}{\left( {}^{{{l}_{1}}}\!\!\diagup\!\!{}_{{{A}_{1}}{{k}_{1}}}\; \right)}+\frac{{{T}_{L}}}{\left( {}^{{{l}_{2}}}\!\!\diagup\!\!{}_{{{A}_{2}}{{k}_{2}}}\; \right)}=T\left( \frac{1}{\left( {}^{{{l}_{1}}}\!\!\diagup\!\!{}_{{{A}_{1}}{{k}_{1}}}\; \right)}+\frac{1}{\left( {}^{{{l}_{2}}}\!\!\diagup\!\!{}_{{{A}_{2}}{{k}_{2}}}\; \right)} \right)$

$\displaystyle \Rightarrow \frac{{{A}_{1}}{{k}_{1}}{{T}_{H}}}{{{l}_{1}}}+\frac{{{A}_{2}}{{k}_{2}}{{T}_{L}}}{{{l}_{2}}}=T\left( \frac{{{A}_{1}}{{k}_{1}}}{{{l}_{1}}}+\frac{{{A}_{2}}{{k}_{2}}}{{{l}_{2}}} \right)$

$\displaystyle \Rightarrow {{A}_{1}}{{k}_{1}}{{T}_{H}}{{l}_{2}}+{{A}_{2}}{{k}_{2}}{{T}_{L}}{{l}_{1}}=T\left( {{A}_{1}}{{k}_{1}}{{l}_{2}}+{{A}_{2}}{{k}_{2}}{{l}_{1}} \right)$

$\displaystyle \Rightarrow T=\frac{{{A}_{1}}{{k}_{1}}{{T}_{H}}{{l}_{2}}+{{A}_{2}}{{k}_{2}}{{T}_{L}}{{l}_{1}}}{{{A}_{1}}{{k}_{1}}{{l}_{2}}+{{A}_{2}}{{k}_{2}}{{l}_{1}}}$ …..(7)

Equivalent thermal conductivity of slabs in series

$\displaystyle {{R}_{eq}}={{R}_{1}}+{{R}_{2}}$

$\displaystyle \Rightarrow \frac{{{l}_{1}}+{{l}_{2}}}{A{{k}_{eq}}}=\frac{{{l}_{1}}}{A{{k}_{1}}}+\frac{{{l}_{2}}}{A{{k}_{2}}}$

Equivalent thermal conductivity of the system is given by :

$\displaystyle {{k}_{eq}}=\frac{{{l}_{1}}+{{l}_{2}}}{\frac{{{l}_{1}}}{{{k}_{1}}}+\frac{{{l}_{2}}}{{{k}_{2}}}}=\frac{\sum{{{l}_{i}}}}{\sum{\frac{{{l}_{i}}}{{{k}_{i}}}}}$ …..(8)

2. Parallel Thermal Resistance: In a parallel arrangement, multiple slabs are placed side by side, and heat flow can take multiple paths through them. The total thermal resistance of the parallel combination is determined by the reciprocal of the sum of the reciprocals of the individual thermal resistances of each slab.

Consider two slabs of length l are held between the same heat reservoirs, their thermal conductivities are k₁ and k₂ and cross-sectional areas are A₁ and A₂.

Thermal current through the slab 1 :

$\displaystyle {{\left( \frac{Q}{t} \right)}_{1}}=\frac{({{T}_{H}}-{{T}_{L}})}{{{R}_{1}}}$ …..(9)

& thermal current through the slab 2 :

$\displaystyle {{\left( \frac{Q}{t} \right)}_{2}}=\frac{({{T}_{H}}-{{T}_{L}})}{{{R}_{2}}}$ …..(10)

Adding eq. (9) and (10) we get net heat current from the hot to cold reservoir,

$\displaystyle {{\left( \frac{Q}{t} \right)}_{net}}={{\left( \frac{Q}{t} \right)}_{1}}+{{\left( \frac{Q}{t} \right)}_{2}}=\frac{({{T}_{H}}-{{T}_{L}})}{{{R}_{1}}}+\frac{({{T}_{H}}-{{T}_{L}})}{{{R}_{2}}}$

$\displaystyle \Rightarrow {{\left( \frac{Q}{t} \right)}_{net}}=({{T}_{H}}-{{T}_{L}})\left[ \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right]$ …..(11)

If the equivalent thermal resistance of the parallel combination of above sabs is R_Parallel, then

$\displaystyle {{\left( \frac{Q}{t} \right)}_{net}}=\frac{({{T}_{H}}-{{T}_{L}})}{{{R}_{Parallel}}}$ …..(12)

Comparing equations (11) and (12), we get

$\displaystyle \frac{1}{{{R}_{Parallel}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}$ …..(13)

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

$\displaystyle \frac{1}{{{R}_{Parallel}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}+.....$ …..(14)

Equivalent thermal conductivity of slabs in parallel

$\displaystyle \frac{1}{{{R}_{eq}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}$

$\displaystyle \Rightarrow \frac{\left( {{A}_{1}}+{{A}_{2}} \right){{k}_{eq}}}{l}=\frac{{{A}_{1}}{{k}_{1}}}{l}+\frac{{{A}_{2}}{{k}_{2}}}{l}$

Equivalent thermal conductivity of the system is given by :

$\displaystyle {{k}_{eq}}=\frac{{{A}_{1}}{{k}_{1}}+{{A}_{2}}{{k}_{2}}}{{{A}_{1}}+{{A}_{2}}}=\frac{\sum{{{A}_{i}}{{k}_{i}}}}{\sum{{{A}_{i}}}}$ …..(15)

Example 2.

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 of 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 k₁ and that of brick is (k₂ = 5k₁). 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.

Solution:

If the interface area is A then the thermal resistance of teak wood,

$\displaystyle {{R}_{1}}=\frac{{{L}_{1}}}{A{{k}_{1}}}$

Similarly thermal resistance of brick wall,

$\displaystyle {{R}_{2}}=\frac{{{L}_{2}}}{A{{k}_{2}}}=\frac{5{{L}_{1}}}{A5{{k}_{1}}}=\frac{{{L}_{1}}}{A{{k}_{1}}}={{R}_{1}}$

If we consider thermal resistance of the each of unknown layer as R, then the above wall can be visualized as a series combination of rods :

In steady state thermal current through each wall is same, hence

$\displaystyle \frac{25-20}{{{R}_{1}}}=\frac{20-{{T}_{3}}}{R}=\frac{{{T}_{3}}-{{T}_{4}}}{R}=\frac{{{T}_{4}}+20}{{{R}_{2}}}$

$\displaystyle \Rightarrow \frac{25-20}{{{R}_{1}}}=\frac{{{T}_{4}}+20}{{{R}_{2}}}$

As R₁ = R₂,

$\displaystyle \Rightarrow 25-20={{T}_{4}}+20$

∴ T₄ = –15ºC

Now

$\displaystyle \frac{20-{{T}_{3}}}{R}=\frac{{{T}_{3}}-{{T}_{4}}}{R}$

$\displaystyle \Rightarrow 20-{{T}_{3}}={{T}_{3}}-{{T}_{4}}$

$\displaystyle \Rightarrow 20-{{T}_{3}}={{T}_{3}}+15$

$\displaystyle {{T}_{3}}=\frac{20-15}{2}$

∴ T₃ = 2.5ºC

Example 3.

In example 2, k₁ = 0.125 W/m–ºC, k₂ = 5k₁ = 0.625 W/m–ºC and thermal conductivity of the unknown material is k = 0.25 W/mºC. L₁ = 4cm, L₂ = 5L₁ = 20cm. If the house consists of a single room of total wall area of 100 m², then find the power of the electric heater being used in the room and the thermal resistance of unknown material.

Solution:

1^st Method :

$\displaystyle {{R}_{1}}={{R}_{2}}=\frac{4\times {{10}^{-2}}}{0.125\times 100}=32\times {{10}^{-4}}{}^{\circ }C/W$

Now

$\displaystyle \frac{25-20}{{{R}_{1}}}=\frac{20-2.5}{R}$

$\displaystyle \Rightarrow \frac{5}{32\times {{10}^{-4}}}=\frac{17.5}{R}$

⇒ R = 112 × 10^–4 ºC/W

The equivalent thermal resistance of the entire wall, R_net = R₁ + R₂ + 2R= 288 × 10^–4 ºC/W

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

$\displaystyle {{\left( \frac{Q}{t} \right)}_{net}}=\frac{({{T}_{H}}-{{T}_{L}})}{{{R}_{net}}}=\frac{25-(-20)}{288\times {{10}^{-4}}}=\frac{45\times {{10}^{4}}}{288}Watt$

$\displaystyle {{\left( \frac{Q}{t} \right)}_{net}}=1.56kW$

Hence the heater must supply 1.56 kW to compensate for the outflow of heat.

2^nd Method :

As in steady state thermal current through each wall is same, hence

$\displaystyle {{\left( \frac{Q}{t} \right)}_{net}}=\frac{{{T}_{1}}-{{T}_{2}}}{{{R}_{1}}}=\frac{25-20}{32\times {{10}^{-4}}}=1.56kW$

Finding resistance of unknown material is same as done above in 1^st Method.

Example 4.

Two rods A and B are of equal length. Each rod has its ends at temperature T₁ and T₂ (T₁ > T₂). What is the condition that will ensure equal rates of heat flow through the rods A and B?

Solution:

For the rate of heat flow to be equal,

$\displaystyle {{\left( \frac{Q}{t} \right)}_{1}}={{\left( \frac{Q}{t} \right)}_{2}}$

$\displaystyle \Rightarrow \frac{({{T}_{1}}-{{T}_{2}})}{{{R}_{1}}}=\frac{({{T}_{1}}-{{T}_{2}})}{{{R}_{2}}}$

$\displaystyle \Rightarrow {{R}_{1}}={{R}_{2}}$

$\displaystyle \Rightarrow \frac{l}{{{A}_{1}}{{k}_{1}}}=\frac{l}{{{A}_{1}}{{k}_{1}}}$

$\displaystyle \Rightarrow {{A}_{1}}{{k}_{1}}={{A}_{1}}{{k}_{1}}$

This is the required condition.

Example 5.

A copper rod 2 m long has a circular cross-section of radius 1 cm. One end is kept at 100 ºC and the other at 0 ºC . The surface is insulated so that negligible heat is lost through the surface. In steady state, find

(a) the thermal resistance of the bar

(b) the thermal current H = dQ/dt

(d) the temperature at a distance 25 cm from the hot end.

Thermal conductivity of copper is 401 W /m−K.

Solution:

(a) Thermal resistance of the bar,

$\displaystyle R=\frac{l}{Ak}=\frac{2}{\pi {{({{10}^{-2}})}^{2}}\times 401}=15.9K/W$

(b) Thermal current

$\displaystyle H=\frac{dQ}{dt}=\frac{{{T}_{H}}-{{T}_{L}}}{R}=\frac{100-0}{15.9}=6.3W$

$\displaystyle \frac{dT}{dx}=\frac{0-100}{2}=-50{}^{\circ }C/m$

(d) Let θ be the temperature at 25 cm from the hot end, then

$\displaystyle \frac{dT}{dx}=\frac{\theta -100}{0.25}=-50$

$\displaystyle \Rightarrow \theta -100=-12.5$

$\displaystyle \Rightarrow \theta =87.5{}^{\circ }C$

Example 6.

A double-pane window consists of two glass sheets each of area 1 m² and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state, the room glass interface and the glass outdoor interface are at constant temperature of 27 ºC and 0 ºC respectively. Find the rate of heat flow through the window pane and the temperature of junction planes. (Given, k_glass = 0.8 Wm⁻¹K⁻¹, k_air = 0.08 Wm⁻¹K⁻¹)

Solution:

Total thermal resistance,

$\displaystyle R=2\times \frac{0.01}{0.8\times 1}+\frac{0.05}{0.08\times 1}=0.65K/W$

Hence heat current,

$\displaystyle H=\frac{dQ}{dt}=\frac{\Delta T}{R}=\frac{27-0}{0.65}=41.5W$

Now temperatures of junction planes,

$\displaystyle \frac{27-{{T}_{1}}}{{{R}_{glass}}}=41.5$

$\displaystyle \Rightarrow \frac{27-{{T}_{1}}}{\left[ 0.01/0.8\times 1 \right]}=41.5$

$\displaystyle \Rightarrow {{T}_{1}}=26.48{}^{\circ }C$

Similarly,

$\displaystyle \Rightarrow \frac{{{T}_{2}}-0}{\left[ 0.01/0.8\times 1 \right]}=41.5$

$\displaystyle \Rightarrow {{T}_{2}}=0.52{}^{\circ }C$

Example 7.

Three rods of the same dimensions have thermal conductivities 3k, 2k and k. They are arranged as shown, with their ends at 100 ºC, 50 ºC and 0 ºC. The temperature of their junction is

(a) 75 ºC

(b) 200/3 ºC

(d) 100/3 ºC

Solution:

Let the temperature of junction point is T. As we know that total incoming of heat is equal to total outgoing heat, so applying Kirchhoff’s current law:

$\displaystyle \frac{T-100}{{}^{l}\!\!\diagup\!\!{}_{A3k}\;}+\frac{T-50}{{}^{l}\!\!\diagup\!\!{}_{A2k}\;}+\frac{T-0}{{}^{l}\!\!\diagup\!\!{}_{Ak}\;}=0$

$\displaystyle \Rightarrow 3\left( T-100 \right)+2\left( T-50 \right)+\left( T-0 \right)=0$

$\displaystyle \Rightarrow 3T-300+2T-100+T=0$

$\displaystyle \Rightarrow 6T=400$

$\displaystyle \Rightarrow T=\frac{200}{3}{}^{\circ }C$

Example 8.

Three cylindrical rods A, B and C of equal lengths and equal diameters are joined in series as shown in figure. Their thermal conductivities are 2k, k and 0.5k, respectively. In steady state, if the free ends of rods A and C are at 100 ºC and 0 ºC, respectively, calculate the temperature at the two junction points. What will be the equivalent thermal conductivity? Assume negligible loss of heat by radiation through the curved surface.

Solution:

Let “a” bet the area of cross-section and “l” be the length of each rod. As the rods are connected in series, so equivalent resistance of the combination,

$\displaystyle {{R}_{eq}}={{R}_{A}}+{{R}_{B}}+{{R}_{C}}$

$\displaystyle {{R}_{eq}}=\frac{l}{2ak}+\frac{l}{ak}+\frac{l}{0.5ak}=\frac{7l}{2ak}$ …..(16)

Also if k_eq is equivalent thermal thermal conductivity, then

$\displaystyle {{R}_{eq}}=\frac{{{l}_{total}}}{a{{k}_{eq}}}=\frac{3l}{a{{k}_{eq}}}$ …..(17)

Comparing equations (16) and (17), we get

$\displaystyle \frac{3l}{a{{k}_{eq}}}=\frac{7l}{2ak}$

Hence equivalent thermal conductivity,

$\displaystyle {{k}_{eq}}=\frac{6k}{7}$

Now heat current,

$\displaystyle \frac{dQ}{dt}=\frac{\left( 100-0 \right)}{{{R}_{eq}}}=\frac{100}{{}^{7l}\!\!\diagup\!\!{}_{2ak}\;}=\frac{200ak}{7l}$ …..(18)

Also

$\displaystyle \frac{dQ}{dt}=\frac{100-{{T}_{AB}}}{{{R}_{A}}}=\frac{100-{{T}_{AB}}}{{}^{l}\!\!\diagup\!\!{}_{2ak}\;}=\frac{2ak\left( 100-{{T}_{AB}} \right)}{l}$ …..(19)

From equations (18) and (19),

$\displaystyle \frac{2ak\left( 100-{{T}_{AB}} \right)}{l}=\frac{200ak}{7l}$

$\displaystyle 7\left( 100-{{T}_{AB}} \right)=100$

$\displaystyle \Rightarrow {{T}_{AB}}=\frac{600}{7}{}^{\circ }C$

Again

$\displaystyle \frac{dQ}{dt}=\frac{200ak}{7l}=\frac{\left( {{T}_{BC}}-0 \right)a\times 0.5k}{l}$

$\displaystyle \Rightarrow \frac{200}{7}=\frac{\left( {{T}_{BC}}-0 \right)}{2}$

$\displaystyle \Rightarrow {{T}_{BC}}=\frac{400}{7}{}^{\circ }C$

Hence equivalent thermal conductivity is $\displaystyle {{k}_{eq}}=\frac{6k}{7}$ and temperature of junction points are $\displaystyle {{T}_{AB}}=\frac{600}{7}{}^{\circ }C$ , $\displaystyle {{T}_{BC}}=\frac{400}{7}{}^{\circ }C$ .

Example 9.

Draw T v/s x graph and dT/dx v/s x graph for above example 8.

Solution :

T v/s x graph :-

For rod A

$\displaystyle {{\left( \frac{dT}{dx} \right)}_{A}}=-\left( \frac{100-{{T}_{AB}}}{l} \right)=-\left( \frac{100-{}^{600}\!\!\diagup\!\!{}_{7}\;}{l} \right)=-\frac{100}{7l}$

For rod B

$\displaystyle {{\left( \frac{dT}{dx} \right)}_{B}}=-\left( \frac{{{T}_{AB}}-{{T}_{BC}}}{l} \right)=-\left( \frac{{}^{600}\!\!\diagup\!\!{}_{7}\;-{}^{400}\!\!\diagup\!\!{}_{7}\;}{l} \right)=-\frac{200}{7l}$

For rod C

$\displaystyle {{\left( \frac{dT}{dx} \right)}_{C}}=-\left( \frac{{{T}_{BC}}-0}{l} \right)=-\left( \frac{{}^{400}\!\!\diagup\!\!{}_{7}\;-0}{l} \right)=-\frac{400}{7l}$

dT/dx v/s x graph :-

Example 10.

Five rods of the same dimensions are arranged as shown.

They have thermal conductivities k₁, k₂, k₃, k₄ and k₅. When points A and B are maintained at different temperatures, no heat flows through the central rod. It follows that :

(A) k₁= k₄and k₂= k₃

(B) k₁/ k₄= k₂/ k₃

(D) k₁ k₂= k₃ k₄

Solution:

As no heat current flows through the central rod, we can apply the concept of balanced wheat-stone bridge which is mathematically given as :

$\displaystyle \frac{{{R}_{1}}}{{{R}_{3}}}=\frac{{{R}_{2}}}{{{R}_{4}}}$

$\displaystyle {{R}_{1}}{{R}_{4}}={{R}_{3}}{{R}_{2}}$

$\displaystyle \Rightarrow \frac{l}{A{{k}_{1}}}\times \frac{l}{A{{k}_{4}}}=\frac{l}{A{{k}_{3}}}\times \frac{l}{A{{k}_{2}}}$

$\displaystyle \Rightarrow {{k}_{1}}{{k}_{4}}={{k}_{2}}{{k}_{3}}$

Hence correct option is (C).

Example 11.

The only possibility of heat flow in a thermos flask is through its cork which is 75 cm² in area and 5 cm thick. Its thermal conductivity is 0.0075 cal/cm s ^∘C. The outside temperature is 40 ^∘C and the latent heat of ice is 80 cal/g. Time taken by 500 g of ice at 0 ^∘C in the flask to melt into water at 0 ^∘C is