Topic Covered



We have seen that heat is energy transfer from one system to another or from one part of a system to another part, arising due to temperature difference. What are the different ways by which this energy transfer takes place? There are three distinct modes of heat transfer : conduction, convection and radiation (Fig. 11.13).


Conduction is the mechanism of transfer of heat between two adjacent parts of a body because of their temperature difference.

Suppose one end of a metallic rod is put in a flame, the other end of the rod will soon be so hot that you cannot hold it by your bare hands. Here heat transfer takes place by conduction from the hot end of the rod through its different parts to the other end. Gases are poor thermal conductors while liquids have conductivities intermediate between solids and gases.

Heat conduction may be described quantitatively as the time rate of heat flow in a material for a given temperature difference. Consider a metallic bar of length L and uniform cross section A with its two ends maintained at different temperatures.

This can be done, for example, by putting the ends in thermal contact with large reservoirs at temperatures, say, `T_C` and `T_D` respectively (Fig. 11.14).

Let us assume the ideal condition that the sides of the bar are fully insulated so that no heat is exchanged between the sides and the surroundings.

After sometime, a steady state is reached; the temperature of the bar decreases uniformly with distance from `color{orange}(T_C` to `T_D; (T_C>T_D)).` The reservoir at C supplies heat at a constant rate, which transfers through the bar and is given out at the same rate to the reservoir at D.

It is found experimentally that in this steady state, the rate of flow of heat (or heat current) H is proportional to the temperature difference `color{navy}((T_C – T_D))` and the area of cross section A and is inversely proportional to the length L :

`color{orange}(H = KA (T_C-T_D)/L.)` ..........(11.14)

The constant of proportionality K is called the thermal conductivity of the material. The greater the value of K for a material, the more rapidly will it conduct heat. The SI unit of K is `color{green}(J S^(–1) m^(–1) K^(–1)` or `W m^(–1) K^(–1.))` The thermal conductivities of various substances are listed in Table 11.5.

These values vary slightly with temperature, but can be considered to be constant over a normal temperature range.
Compare the relatively large thermal conductivities of the good thermal conductors, the metals, with the relatively small thermal conductivities of some good thermal insulators, such as wood and glass wool.

You may have noticed that some cooking pots have copper coating on the bottom. Being a good conductor of heat, copper promotes the distribution of heat over the bottom of a pot for uniform cooking.

Plastic foams, on the other hand, are good insulators, mainly because they contain pockets of air. Recall that gases are poor conductors, and note the low thermal conductivity of air in the Table 11.6

Heat retention and transfer are important in many other applications. Houses made of concrete roofs get very hot during summer days, because thermal conductivity of concrete (though much smaller than that of a metal) is still not small enough.

Therefore, people usually prefer to give a layer of earth or foam insulation on the ceiling so that heat transfer is prohibited and keeps the room cooler. In some situations, heat transfer is critical.

In a nuclear reactor, for example, elaborate heat transfer systems need to be installed so that the enormous energy produced by nuclear fission in the core transits out sufficiently fast, thus preventing the core from overheating.

Q 3210712619

What is the temperature of the steel-copper junction in the steady state of the system shown in Fig. 11.15. Length of the steel rod `= 15.0 cm,` length of the copper rod `= 10.0 cm`, temperature of the furnace `= 300 °C`, temperature of the other end `= 0 °C.` The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel `= 50.2 J s^(–1) m^(–1)K^(–1);` and of copper `= 385 J s^(–1)m^(–1)K^(–1)).`
Class 11 Chapter 11 Example 6

The insulating material around the rods reduces heat loss from the sides of the rods. Therefore, heat flows only along the length of the rods. Consider any cross section of the rod. In the steady state, heat flowing into the element must equal the heat flowing out of it; otherwise there would be a net gain or loss of heat by the element and its temperature would not be steady. Thus in the steady state, rate of heat flowing across a cross section of the rod is the same at every point along the length of the combined steel-copper rod. Let T be the temperature of the steel-copper junction in the steady state. Then,

`(K_1A_1 \ \(300- T))/(L_1)= (K_2A_2 (T-0))/(L_2)`
where 1 and 2 refer to the steel and copper rod respectively. For `A_1 = 2 A_2, L_1 = 15.0 cm,`
`L_2 = 10.0 cm, K_1 = 50.2 J s^(–1) m^(–1) K ^(–1), K_2 = 385 J`
`s^(–1) m^(–1) K^(–1)`, we have

`(50.2xx2(300-T))/(15) = (385T)/(10)`
which gives `T = 44.4 °C`
Q 3260812715

An iron bar `(L_1 = 0.1 m, A_1 = 0.02 m^2, K_1 = 79 W m^(–1) K^(–1)` and a brass bar `(L_2 = 0.1 m, A_2 = 0.02 m^2, K_2 = 109 W m–1K–1)` are soldered end to end as shown in Fig. 11.16. The free ends of the iron bar and brass bar are maintained at 373 K and 273 K respectively. Obtain expressions for and hence compute (i) the temperature of the junction of the two bars, (ii) the equivalent thermal conductivity of the compound bar, and (iii) the heat current through the compound bar.
Class 11 Chapter 11 Example 7

Given, `L_1 = L_2= L = 0.1 m, A_1 = A_2= A= 0.02 m^2`
`K_1 = 79 W m^(–1) K^(–1,) K_2 = 109 W m^(–1) K^(–1,)`
`T_1 = 373 K, and T_2 = 273 K.`

Under steady state condition, the heat current `(H_1)` through iron bar is equal to the heat current `(H_2)` through brass bar.
So, `H = H_1 = H_2`

`=(K_1A_1 \ \ (T_1-T_0))/(L_1) = (K_2A_2(T_0-T_2))/(L_2)`
For `A_1 = A_2 = A` and `L_1 = L_2 = L,` this equation leads to `K_1 (T_1 – T_0) = K_2 (T_0 – T_2)`
Thus the junction temperature `T_0` of the two bars is

`T_0 = (K_1T_1 + K_2T_2)/(K_1 + K_2)`
Using this equation, the heat current H through either bar is

`H = (K_1A \ \T_1-T_0)/(L) = (K_2A(T_0-T_2))/L`

`(K_1K_2)/(K_1\ \ K_2) (A \ \ T_1-T_0)/L =(A \ \ T_1-T_2)/(L 1/(K_1) \ \ 1/(K_2)`
Using these equations, the heat current H′ through the compound bar of length `L_1 + L_2 = 2L` and the equivalent thermal conductivity K′, of the compound bar are given by

`(J \ \ K \ \A \ \ (T_1 - T_2))/(2L)H`
`K (2\ \ K_1\ \ K_2)/(K_1 \ \K_2)`

`(i) T_0 = (K_1T_1 + K_2T_2)/(K_1 + K_2)`

`(79Wm^(-1)K^(-1) \ \373K \ \ \ \ 109Wm^(-1)K^(-1) \ \ \ 273K)/(79 Wm^(-1)K^(-1) \ \ \109 Wm^(-1)K^(-1))`

`= 316K`

`(ii)K' = (2K_1 \ \K_2)/(K_1 + K_2)`

`= (2xx(79Wm^(-1)k^(-1)) + (109Wm^(-1)K^(-1)))/(79Wm^(-1)K^(-1) + 109Wm^(-1)K^(-1))`
`= 91.6Wm^(-1)K^(-1)`

`(iii) H'= H = (K \ \A \ \T_1-T_2)/(2L)`

`(91.6Wm^(-1)K^(_1) xx0.02m^2xx 373K-273K)/(2xx0.1m)`
`= 916.1W`


Convection is a mode of heat transfer by actual motion of matter. It is possible only in fluids. Convection can be natural or forced.

In natural convection, gravity plays an important part. When a fluid is heated from below, the hot part expands and, therefore, becomes less dense.

Because of buoyancy, it rises and the upper colder part replaces it. This again gets heated, rises up and is replaced by the colder part of the fluid. The process goes on. This mode of heat transfer is evidently different from conduction.

Convection involves bulk transport of different parts of the fluid. In forced convection, material is forced to move by a pump or by some other physical means. The common examples of forced convection systems are forced-air heating systems in home, the human circulatory system, and the cooling system of an automobile engine.

In the human body, the heart acts as the pump that circulates blood through different parts of the body, transferring heat by forced convection and maintaining it at a uniform temperature.

Natural convection is responsible for many familiar phenomena. During the day, the ground heats up more quickly than large bodies of water do. This occurs both because the water has a greater specific heat and because mixing currents disperse the absorbed heat throughout the great volume of water.

The air in contact with the warm ground is heated by conduction. It expands, becoming less dense than the surrounding cooler air. As a result, the warm air rises (air currents) and other air moves (winds) to fill the space-creating a sea breeze near a large body of water.

Cooler air descends, and a thermal convection cycle is set up, which transfers heat away from the land. At night, the ground loses its heat more quickly, and the water surface is warmer than the land. As a result, the cycle is reveresed (Fig. 11.17).

The other example of natural convection is the steady surface wind on the earth blowing in from north-east towards the equator, the so called trade wind.

A resonable explanation is as follows : the equatorial and polar regions of the earth receive unequal solar heat. Air at the earth’s surface near the equator is hot while the air in the upper atmosphere of the poles is cool. In the absence of any other factor, a convection current would be set up, with the air at the equatorial surface rising and moving out towards the poles, descending and streaming in towards the equator.

The rotation of the earth, however, modifies this convection current. Because of this, air close to the equator has an eastward speed of 1600 km/h, while it is zero close to the poles. As a result, the air descends not at the poles but at 30° N (North) latitude and returns to the equator. This is called trade wind.


Conduction and convection require some material as a transport medium. These modes of heat transfer cannot operate between bodies separated by a distance in vacuum.

But the earth does receive heat from the sun across a huge distance and we quickly feel the warmth of the fire nearby even though air conducts poorly and before convection can set in. The third mechanism for heat transfer needs no medium; it is called radiation and the energy so radiated by electromagnetic waves is called radiant energy.

In an electromagnetic wave electric and magnetic fields oscillate in space and time. Like any wave, electromagnetic waves can have different wavelengths and can travel in vacuum with the same speed, namely the speed of light i.e., `3 xx 10^8 m s^(–1)` .

You will learn these matters in more details later, but you now know why heat transfer by radiation does not need any medium and why it is so fast. This is how heat is transfered to the earth from the sun through empty space.

All bodies emit radiant energy, whether they are solid, liquid or gases. The electromagnetic radiation emitted by a body by virtue of its temperature like the radiation by a red hot iron or light from a filament lamp is called thermal radiation.

When this thermal radiation falls on other bodies, it is partly reflected and partly absorbed. The amount of heat that a body can absorb by radiation depends on the colour of the body. We find that black bodies absorb and emit radiant energy better than bodies of lighter colours.

This fact finds many applications in our daily life. We wear white or light coloured clothes in summer so that they absorb the least heat from the sun. However, during winter, we use dark coloured clothes which absorb heat from the sun and keep our body warm.

The bottoms of the utensils for cooking food are blackened so that they absorb maximum heat from the fire and give it to the vegetables to be cooked.

Similarly, a Dewar flask or thermos bottle is a device to minimise heat transfer between the contents of the bottle and outside. It consists of a double-walled glass vessel with the inner and outer walls coated with silver. Radiation from the inner wall is reflected back into the contents of the bottle.

The outer wall similarly reflects back any incoming radiation. The space between the walls is evacuted to reduce conduction and convection losses and the flask is supported on an insulator like cork. The device is, therefore, useful for preventing hot contents (like milk) from getting cold, or alternatively to store cold contents (like ice).


We all know that hot water or milk when left on a table begins to cool gradually. Ultimately it attains the temperature of the surroundings. To study how a given body can cool on exchanging heat with its surroundings, let us perform the following activity.

Take some water, say 300 ml, in a calorimeter with a stirrer and cover it with two holed lid. Fix a thermometer through a hole in the lid and make sure that the bulb of thermometer is immersed in the water.

`"Note"` the reading of the thermometer. This reading `T_1` is the temperature of the surroundings. Heat the water kept in the calorimeter till it attains a temperature, say, `40 °C` above room temperature (i.e., temperature of the surroundings). Then stop heating the water by removing the heat source.

Start the stopwatch and note the reading of the thermometer after fixed interval of time, say after every one minute of stirring gently with the stirrer. Continue to note the temperature `(T_2)` of water till it attains a temperature about `5 °C` above that of the surroundings.

Then plot a graph by taking each value of temperature `color{blue}(ΔT = T_2 – T_1)` along y axis and the coresponding value of t along x-axis (Fig. 11.18).

From the graph you will infer how the cooling of hot water depends on the difference of its temperature from that of the surroundings.

You will also notice that initially the rate of cooling is higher and decreases as the temperature of the body falls.
The above activity shows that a hot body loses heat to its surroundings in the form of heat radiation. The rate of loss of heat depends on the difference in temperature between the body and its surroundings. Newton was the first to study, in a systematic manner, the relation between the heat lost by a body in a given enclosure and its temperature.

According to Newton’s law of cooling, the rate of loss of heat, `– dQ//dt` of the body is directly proportional to the difference of temperature `ΔT = (T_2–T_1)` of the body and the surroundings.

The law holds good only for small difference of temperature. Also, the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface. We can write

`color{green}(- (dQ)/(dt) = k (T_2\ \T_1))`....(11.15)

where k is a positive constant depending upon the area and nature of the surface of the body. Suppose a body of mass m and specific heat capacity s is at temperature `T_2.` Let `T_1` be the temperature of the surroundings. If the temperature falls by a small amount d `T_2` in time dt, then the amount of heat lost is

`color{navy}(dQ = ms dT_2)`

:. Rate of loss of heat is given by

`color{orange}((dQ)/(dt) =ms (dT_2)/(dt))`........(11.16)
From Eqs. (11.15) and (11.16) we have

`color{green}(-ms(dT_2)/(dT) = k (T_2-T_1))`

`color{orange}((dT_2)/(T_2_T_1)=-k/ms dt=-Kdt)`....(11.17)

where `K = k//m s`

On integrating, `color{red}(loge (T_2 – T_1) = – K t + c)`........(11.18)

or `color{orange}(T_2 = T_1 + C′ e^(–Kt); \ \ \ \ where C′ = e^c)`..... (11.19)

Equation (11.19) enables you to calculate the time of cooling of a body through a particular range of temperature.

For small temperature differences, the rate of cooling, due to conduction, convection, and radiation combined, is proportional to the difference in temperature. It is a valid approximation in the transfer of heat from a radiator to a room, the loss of heat through the wall of a room, or the cooling of a cup of tea on the table.

Newton’s law of cooling can be verified with the help of the experimental set-up shown in Fig. 11.19(a). The set-up consists of a double walled vessel (V) containing water in between the two walls.

A copper calorimeter (C) containing hot water is placed inside the double walled vessel. Two thermometers through the corks are used to note the temperatures `T_2` of water in calorimeter and `T_1` of hot water in between the double walls respectively.

Temperature of hot water in the calorimeter is noted after equal intervals of time. A graph is plotted between loge `(T_2–T_1)` and time (t). The nature of the graph is observed to be a straight line having a negative slope as shown in Fig. 11.19(b). This is in support of Eq. (11.18).
Q 3260012815

A pan filled with hot food cools from `94 °C` to `86 °C` in 2 minutes when the room temperature is at `20 °C`. How long will it take to cool from `71 °C` to `69 °C`?
Class 11 Chapter 11 Example 8

The average temperature of `94 °C` and `86 °C` is `90 °C,` which is `70 °C` above the room temperature. Under these conditions the pan cools `8 °C` in 2 minutes
Using Eq. (11.17), we have

`("Change in temperature")/("Time") = K DeltaT`

`(8^oC)/(2 min) = K (70^oC)`

The average of `69 °C` and `71 °C` is `70 °C,` which is `50 °C` above room temperature. K is the same for this situation as for the original.

`(2^oC)/(Time)= K(50^oC)`
When we divide above two equations, we have

`(8 \ \C //2min)/(2C//time) = (K(70C))/(K(50C)`
Time `= 0.7 min`
`= 42 s`