Thermal Properties of Matter · Question 32

The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T1 and T2 (T1 > T2). The rate of heat transfer, dQdt through the rod in a steady state is given by :

1. dQdt = k(T1 - T2)LA
2. dQdt = kLA(T1 - T2)
3. dQdt = kA(T1 - T2)L
4. dQdt = kL(T1 - T2)A

On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39 and 239 respectively. What will be the temperature on the new scale, corresponding to a temperature of 39 on the Celsius scale ?

1. 200
2. 139
3. 78
4. 117

Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at temperature t C, the power received by a unit surface, (normal to the incident rays) at a distance R from the centre of the sun is where is the Stefan's constant.

1. r2 (t+273)44 2
2. 16 2r2 4R2
3. r2 (t+273)4R2
4. 4 2 4R2

A black body is at 727 . It emits energy at a rate which is proportional to

1. (1000)4
2. (1000)2
3. (727)4
4. (727)2

A black body at 1227 emits radiations with maximum intensity at a wavelength of 5000 A. If the temperature of the body is increased by 1000 , the maximum intensity will be observed at

1. 3000 A
2. 4000 A
3. 5000 A
4. 6000 A

Which of the following rods, (given radius r and length l) each made of the same material and whose ends are maintained at the same temperature will conduct most heat?

1. r = r0, l = l0
2. r = 2r0, l = l0
3. r = r0, l = 2l0
4. r = 2r0, l = 2l0

If m denotes the wavelength at which the radiative emission from a black body at a temperature T K is maximum, then

1. m T4
2. m is independent of T
3. m T
4. m T-1

Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities K and 2K, respectively. The equivalent thermal conductivity of the slab is

1. 23K
2. 2K
3. 3K
4. 43K

Unit of Stefan's constant is

1. watt m2 K4
2. watt m2/K4
3. watt/m2 K
4. watt/m2K4

Consider two rods of same length and different specific heats (S1, S2), conductivities (K1, K2) and area of cross-sections (A1, A2) and both having temperatures T1 and T2 at their ends. If rate of loss of heat due to conduction is equal, then

1. K1A1 = K2A2
2. K1A1S1 = K2A2S2
3. K2A1 = K1A2
4. K2A1S2 = K1A2S1