Questions OBJECTIVE - I and Answer Heat Transfer HC Verma Part II

Q#1

The thermal conductivity of a rod depends on

(a) length

(b) mass

(c) area of cross-section

(d) the material of the rod.   

 

Answer:  (d)    

Explanation:  Thermal conductivity is a property of the material like the electrical conductivity.

Q#2

In a room containing air, heat can go from one place to another

(a) by conduction only

(b) by convection only

(c) by radiation only

(d) by all the three modes.   

Answer:  (d)    

Explanation:  Heat can go through the air by conduction as well as convection. Also, heat is radiated by every object at all the temperatures and in this mode, no medium is required. So in the room heat can go from one to another by all the three means.

Q#3

A solid at temperature T₁ is kept in an evacuated chamber at temperature T₂ > T₁. The rate of increase in temperature of the body is proportional to

(a) T₂ - T₁

(b) T₂² - T₁²

(c) T₂³ - T₁³  

(d) T₂⁴ - T₁⁴      

Answer:  (d)    

Explanation:  Since the chamber is evacuated only mode of heat transfer will be by radiation. Thermal radiation emitted by the body is

u₁ = eσAT₁⁴

Where e = emissivity of the body, A = surface area of the body and σ = Stefan-Boltzmann constant.

And the thermal radiation received by the body is

u₂ = eσAT₂⁴ 

Since T₂ > T₁, net heat received by the body is

u = u₂ - u₁ = eσA(T₂⁴ - T₁⁴)

The increase in temperature of the body

T = u/ms

where m is mass and s is specific heat. 

So, T ∝ u

 and u ∝ T₂⁴ - T₁⁴ 

Thus T ∝  T₂⁴ - T₁⁴  

Q#5

The thermal radiation emitted by a body is proportional to Tⁿ when T is its absolute temperature. The value of n is exactly 4 for

(a) a blackbody

(b) all bodies

(c) bodies painted black only

(d) polished bodies only.   

Answer:  (b)    

Explanation:  Thermal radiation emitted by all bodies is proportional to T⁴ though it is maximum by a blackbody. For other bodies, it is less by a factor e (called emissivity of the body) which has a value between 0 and 1. For a blackbody e = 1.   

Q#5

Two bodies A and B having equal surface areas are maintained at temperatures 10°C and 20°C. The thermal radiation emitted in a given time by A and B are in the ratio

(a) 1:1.15

(b) 1:2

(c) 1:4

(d) 1:16.   

 

Answer: (a)    

Explanation:  Here the absolute temperature of A, T₁ = 273 + 10 = 283 K and that of B is T₂ = 273 + 20 = 293 K. Since the thermal radiation emitted by a body is proportional to T⁴, the ratio of thermal radiation emitted in a given time for A and B will be

= 283⁴ : 293⁴

= 1 : (293/283)⁴

= 1 : 1.15 

Q#6

One end of a metal rod is kept in a furnace. In steady-state, the temperature of the rod

(a) increases 

(b) decreases

(c) remains constant

(d) is non-uniform.  

Answer:  (d)    

Explanation: Since the temperature of the furnace will be greater than the surrounding the heat will flow at a steady rate from a higher temperature to the lower temperature in the steady-state. So the temperature of the rod will be maximum near the furnace and gradually decrease to the other end. Thus the temperature of the rod will be non-uniform.  

Q#7

Newton's law of cooling is a special case of

(a) Wien's displacement law

(b) Kirchoff's law

(c) Stefan's law

(d) Plank's law.      

Answer:  (c)    

Explanation: Newton's law of cooling is a special case of Stefan's law when the temperature difference between the body and the surrounding is small.  The fourth power of the temperature, in this case, is expanded using binomial expression and neglecting the higher powers of ΔT/T to derive Newton's law of cooling.  

Q#8

A hot liquid is kept in a big room. Its temperature is plotted as a function of time. Which of the following curves may represent the plot?  

Answer:  (a)    

Explanation: Since the liquid is hot initially the loss of radiation will be proportional to T₂⁴-T₁⁴. So the slope of the curve will be steep initially. As the difference decreases with time the slope will get mild. And when the difference is very small the loss will be nearly proportional to T₂-T₁ as per Newton's law of cooling. At this time the curve will be nearly straight. So the plotted curve will resemble (a).     

Q#9

A hot liquid is kept in a big room. The logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time. The plot will be very nearly 

(a) a straight line

(b) a circular arc

(c) a parabola

(d) an ellipse.   

Answer:  (a)    

Explanation: According to Newton's law of cooling

dT/dt = -bA(T - T') where T is the temperature of the liquid and T' is the temperature of the room. A is the surface area of the liquid and b is constant. It can be written as

dT/(T - T') = -bAdt

integrating we get

ln(T - T') = - bA*t + C

Clearly, the variation of time t with the logarithm of (T - T') is linear one hence its graph will be a straight line.     

Q#10

A body cools down from 65°C to 60°C in 5 minutes. It will cool down from 60° to 55°C in

(a) 5 minutes

(b) less than 5 minutes

(c) more than 5 minutes

(d) less than or more than 5 minutes depending on whether its mass is more than or less than 1 kg.   

Answer: (c)

Explanation:  Suppose the temperature of the surrounding = T. Average temperature in the first case 

= (65 + 60)/2 = 62.5°C.

The difference in temperature 

= (62.5 - T)°C

In the second case, average temperature = (60 + 55)/2 = 57.5°C.

The difference in temperature now 

= (57.5 - T)°C.

Clearly, (62.5 - T) > (57.5 - T)

According to Newton's cooling law, the rate of cooling is proportional to the temperature difference. Hence in the second case, the rate of cooling will be less than the first case.

Thus, it will take more time to cool from 60°C to 55°C that it takes to cool from 65°C to 60°C.