Questions (MCQ) on Tansmission of Heat

In this section you have to remember the following:

(1) The quantity (Q) of heat conducted through a rod of thermal conductivity ‘K’ and cross section area ‘A’ in time ‘t’ under a temperature gradient dθ/dx is given by

Q = KA(dθ/dx)t

(2) If ‘n’ conductors of the same area of cross section having lengths d₁, d₂, d₃….. d_n and thermal conductivities K₁, K₁, K₃,….K_n are joined in series, the same quantity of heat will flow through them in a given time. The thermal conductivity of such a compound rod is given by K = (d₁+ d₂+ d₃+….. +d_n)/[(d₁/K₁)+ (d₂/K₂)+ (d₃/K₃)+…...+(d_n/K_n)]

If two rods of the same length and the same area of cross section are joined in series, the equivalent thermal conductivity can be obtained from the above general expression as K = 2K₁K₂/ (K₁+K₂)

(3) Stefan’s law: Energy (E) radiated per second from a perfectly black body of surface area ‘A’ is given by E = AσT⁴ where ‘σ’ is Stefan’s constant equal to 5.57×10^–8Wm^–2K^–4.

If the body is not a perfect black body, E = AeσT⁴ where ‘e’ is a dimensionless fraction called emissivity.

In the case of a perfectly black body at temperature T, with surroundings at temperature T_0, the net rate of loss of radiant energy is E = Aσ(T⁴ – T₀⁴), since the body emits energy AσT⁴ and absorbs energy AσT₀⁴ per second in accordance with Kirchhoff's law.

(4) Wein’s displacement law: λ_mT = constant (= 0.29cmK) where λ_m is the wavelength for which the energy radiated is maximum at temperature T. If the constant is taken as 0.29 cmK, the value of λ_m is to be substituted in cm.

(5) Newton’s law of cooling: The rate of cooling is directly proportional to the excess of temperature of the body over the surroundings if the excess of temperature is small compared to the temperature of the body (T) and the temperature (T_S) of the surroundings: dT/dt α (T– T_S)

Now consider the following MCQ which appeared in IIT screening 2002 question paper:

An ideal black body at room temperature is thrown in to a furnace. It is observed that

(a) initially it is the darkest body and at later times the brightest

(b) it is the darkest body at all times

(c) it cannot be distinguished at all times

(d) initially it is the darkest body and at later times it cannot be distinguished

A black body is a good emitter as well as a good absorber. Initially it will absorb energy and hence will appear dark. Once it attains the temperature of the furnace, it emits better than the other parts of the furnace and hence appears the brightest. So the correct option is (a).

Consider the following question:

Two copper spheres A and B having the same emissivity have radii 12 cm and 3 cm respectively . If they are maintained at temperatures 727°C and 1727°C respectively, the ratio of energy radiated by A and B is

(a) 0.032 (b) 0.12 (c) 0.48 (d) 1 (e) 2

Since the energy radiated per second is given by E = AeσT⁴ and the emissivities are equal, we have E₁/E₂ = (A₁/A₂)(T₁/ T₂ )⁴.

Since the radii of A and B are in the ratio 4:1, the surface areas are in the ratio 16:1. Since the temperatures are 1000K and 2000K, the ratio (T₁/ T₂ )⁴ = 1/16. Therefore, E₁/E₂ = 1.

Here is a simple question involving Wein’s displacement law:
On examining the spectrum of a star, it is found that mximum energy is emitted at a wave length of 5800 Ǻ. The surface temperature of the star is
(a) 4500 K (b) 5000 K (c) 5500 K (d) 6000 K (e) 6500 K
This question indicates how one can determine the temperature of a distant star from spectroscopic data. According to Wien’s law, we have λ_mT = 0.29cmK, from which T =0.29/(5800×10^–8) = 5000K. [Note that we have substituted the wave length in cm].

Consider the following question involving heat conduction:

Three identical iron rods are welded together to form the shape of Y. The top ends of the ‘Y’ are maintained at 0°C and the bottom end is maintained at 600°C. The temperature of the junction of the three rods is

(a) 100°C (b) 200°C (c) 250°C (d) 300°C (e) 400°C

The quantity of heat conducted per second through the bottom rod making the ‘Y’ gets divided equally at the junction of the three rods. If ‘θ’ is the temperature of the junction, we have

KA(θ–0)/L = 2KA(600 – θ)/L where K is the thermal conductivity, A is the area of cross section and L is the length of the identical rods.

[ Note that the L.H.S. is the quantity of heat conducted through the lower single rod making the ‘Y’ and the R.H.S. is the sum of the quantities of heat conducted through the upper two rods].

The above equation yields θ = 400°C