Heat Transfer(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

HEAT TRANSFER MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1. A spherical ball A of surface area 20

c m^{2}

is kept at the centre of a hollow spherical shell B of area 80

c m^{2}

. The surface of A and the inner surface of B emit as blackbodies. Assume that the thermal conductivity of the material of B is extremely poor and that of A is very high and that the air between A and B has been pumped out. The heat capacities of A and B are 42 J/

^{\circ}

C and 82 J/

^{\circ}

C respectively. Initially the temperature of A is 100

^{\circ}

C and that of B is 20

^{\circ}

C.Find the rate of change of temperature of A and that of B at this instant.

0.005∘c/s and 0.0001∘c/s
0.05∘c/s and 0.01∘c/s
0.025∘c/s and 0.0050∘c/s
0.03∘c/s and 0.01∘c/s

Discuss Question

Answer: Option D. -> 0.03∘c/s and 0.01∘c/s
:
D
The inner surface of B radiates at a rate of

\frac{Δ Q}{Δ t} = σ A_{B} T_{A}^{4}

,
where

σ \to

stefan's constant
AB

\to

Area
TB

\to

Temperature
Similarly, A radiates

\frac{Δ Q_{2}}{Δ t} = σ A_{B} T_{A}^{4}

Now, all the radiation from inner surface of B goes inside the hollow of the shell, as B is a very poor conductor. So, this radiation falls on A, and A being a blackbody absorbs all the radiation falling on it! And it being a good conductor the heat travels to the centre almost instantly, thus making the temperature uniform throughout the ball. Similarly, all the heat radiated by A is completely absorbed by inner surface of B, and none of it is lost to the surrounding as B is a very poor conductor.
∴ Net heat lost by B = Net heat gained by A =

σ (A_{B} T_{B}^{4} - A_{A} T_{A}^{4})

5.67 \times 10^{- 8} \times (80 \times 293.1 s^{4} - 20 \times 373.1 s^{4}) \times 10^{- 4}

= 1.15 w
Now this is equal to specific heat

\times \frac{d t e m p e r a t u r e}{d T i m e}

of both A and B

∴ \frac{d T_{A}}{d t} = {\frac{1.15}{42}}^{\circ} c / s = {0.03}^{\circ} c / s

\frac{d T_{B}}{d t} = {\frac{1.15}{82}}^{\circ} c / s = {0.01}^{\circ} c / s

Question 2. The temperature of a body falls from

40^{\circ} C

36^{\circ} C

in 5 minutes when placed in a surrounding of constant temperature

16^{\circ} C

. Find the time taken for the temperature of the body to become

32^{\circ} C

8.1 min
6.1 min
5.3 min
5.1 min

Discuss Question

Answer: Option B. -> 6.1 min
:
B
As the temperature differences are small, we can use Newton's law of cooling.

\frac{d θ}{d t} = - k (θ - θ_{0}) \frac{d θ}{θ - θ_{0}} = - k d t

...(i)
Where k is a constant,

θ

is the temperature of the body at time t and

θ_{0}

16^{\circ} C

is the temperature of the surrounding. We have,

\int_{40^{0} c}^{36^{0} c} \frac{d θ}{θ - θ_{0}} = - k (5 m i n) I n \frac{36^{0} c - 16^{0} c}{40^{0} c - 16^{0} c} = - k (5 m i n)

or,

k = - \frac{I n (5 / 6)}{5 m i n}

or,If t be the time required for the temperature to fall from 36⁰C to 32⁰C then by (i),

\int_{36^{0} c}^{32^{0} c} \frac{d θ}{θ - θ_{0}} = - k t I n \frac{32^{0} c - 16^{c}}{36^{0} c - 16^{0}} = - \frac{I n (5 / 6) t}{5 m i n} t = \frac{I n (4 / 5)}{I n (5 / 6)} \times 5 m i n = 6.1 m i n

Question 3. The graph shown in the adjacent diagram, represents the variation of temperature (T) of two bodies, x and y having same surface area, with time (t) due to the emission of radiation. Find the correct relation between the emissivity (

e

) and absorptivity (

a

) power of the two bodies. (IIT JEE 2003)

The Graph Shown In The Adjacent Diagram, Represents The Vari...

ex>ey,ax>ay
ex>ey,ay>ax
ey>ex,ay>ax
ey>ex,ax>ay

Discuss Question

Answer: Option A. -> ex>ey,ax>ay
:
A
From the graph it is clear that initially both the bodies are at same temperature but after that at any instant temperature of body x is less than the temperature of body y. It means body x emits more heat i.e., emissivity of body x is more than body y

∴ e_{x} > e_{y}

and according to Kirchhoff's law good emitter are also good absorber so

a_{x} > a_{y}

Question 4. A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of 0.1gm of ice per second. If the rod is replaced by another with half the length and double the radius of the first and if the thermal conductivity of material of second rod is

\frac{1}{4}

that of first, the rate at which ice melts in gm/sec will be

Discuss Question

Answer: Option C. -> 0.2
:
C

\frac{Q}{t} = \frac{K A Δ θ}{l} \Rightarrow \frac{m L}{t} = \frac{K (π r^{2}) Δ θ}{l}

\Rightarrow

Rate of melting of ice

(\frac{m}{t}) \propto \frac{K r^{2}}{l}

Since for second rod K becomes

\frac{1}{4} t h

r becomes double and length becomes half, so rate of melting will
be twice i.e.

{(\frac{m}{t})}_{2} = 2 {(\frac{m}{t})}_{1} = 2 \times 0.1 = 0.2 g m / s e c .

Question 5. A sphere and a cube of same material and same volume are heated upto same temperature and allowed to cool in the same surroundings. The ratio of the amounts of radiations emitted will be

1 : 1
4π3:1
(π6)13:1
12(4π3)23:1

Discuss Question

Answer: Option C. -> (π6)13:1
:
C

Q = σ A t (T^{4} - T_{0}^{4})

T, T_{0}, σ

and t are same for both bodies then

\frac{Q_{s p h e r e}}{Q_{c u b e}} = \frac{A_{s p h e r e}}{A_{c u b e}} = \frac{4 π r^{2}}{6 a^{2}}

.....(i)
But according to problem, volume of sphere = Volume of cube

\Rightarrow \frac{4}{3} π r^{3} = a^{3} \Rightarrow a = {(\frac{4}{3} π)}^{\frac{1}{3}} r

Substituting the value of a in equation (i) we get

\frac{Q_{s p h e r e}}{Q_{c u b e}} = \frac{4 π r^{2}}{6 a^{2}} = \frac{4 π r^{2}}{6 {{(\frac{4}{3} π)}^{\frac{1}{3}} r}^{2}}

= \frac{4 π r^{2}}{6 {(\frac{4}{3} π)}^{\frac{2}{3}} r^{2}} = {(\frac{π}{6})}^{\frac{1}{3}} : 1

Question 6. Polyurethane is quite good for cold weather for jackets and blankets. However, thermal emergency blankets or space blankets are often made up of something similar to aluminum foil and used by astronauts to avoid hypothermia when stuck in subzero temperature. Why do you think it is a good choice?

Thermal blankets reflect the body heat back, thereby trapping it inside
Thermal blankets absorb heat from cold surroundings
Thermal blankets don't work
None of the above

Discuss Question

Answer: Option A. -> Thermal blankets reflect the body heat back, thereby trapping it inside
:
A
'Aluminum foil like' thermal blankets are designed to reflect most of the radiation, or infrared radiation.
Also, they are shiny! This makes them easy to spot from helicopters etc, in case of rescue operations.
So, all in all, a must have for campers!

Question 7. The earth receives solar radiation at a rate of 8.2 J/

c m^{2}

- minute. Assuming that the sun radiates like a blackbody, calculate the surface temperature of the sun. The angle subtended by the sun on the earth is 0.53

^{\circ}

and the Stefan constant

σ = 5.67 \times 10^{- 8} W / m^{2}

- K4.

6800 K
5800 K
7800 K
7800 K

Discuss Question

Answer: Option B. -> 5800 K
:
B
Let the diameter of the sun be D and its distance from the earth be R. From the question,

\frac{D}{R} \approx 0.53 \times \frac{π}{180}

9.25 \times 10^{- 3}

. ...(i)
The radiation emitted by the surface fo the sun per unit time is
At distance R, this radiation falls on an area

4 π R^{2}

in unit time. The radiation received at the earth's surface per unit time per unit area is, therefore,

Question 8. The adjoining diagram shows the spectral energy density distribution

E_{λ}

of a black body at two different temperatures. If the areas under the curves are in the ratio 16 : 1, find the value of temperature T.

The Adjoining Diagram Shows The Spectral Energy Density Dist...

3000K
4000 K
5000 K
6000 K

Discuss Question

Answer: Option B. -> 4000 K
:
B
Area under curve represents the emissive power of the body

\frac{E_{T}}{E_{2000}} = \frac{A_{T}}{A_{2000}} = \frac{16}{1}

(given) ...(i)
But from Stefan's law E α T4

∴ \frac{E_{T}}{E_{2000}} = {(\frac{T}{2000})}^{4}

...(ii)
From (i) and (ii)

{(\frac{T}{2000})}^{4} = \frac{16}{1}

\Rightarrow \frac{T}{2000} = 2 \Rightarrow T = 4000 k

Question 9. Penguins huddling: To withstand the harsh weather of the Antarctic, emperor penguins huddle in groups. Assume that a penguin is a circular cylinder with a top surface area a =0.26

m^{2}

and height h = 90 cm. Let

P_{r}

be the rate at which an individual penguin radiates energy to the environment (through the top and the sides); thus

N P_{r}

is the rate at which N identical, well-separated penguins radiate. If the penguins huddle closely to form a huddled cylinder with top surface area Na and height h, the cylinder radiates at the rate

P_{h}

. If N = 1000, by what percentage does huddling reduce the total radiation loss?

Discuss Question

Answer: Option D. -> 27%
:
D
Heat loss by individula penguin =

σ e A_{p e n g u i n} (T_{p e n g u i n}^{4} - T_{e n v i r o n m e n t}^{4})

σ e (2 π r_{p e n g u i n} h + 2 π r_{p e n g u i n}^{2}) (T_{p e n g u i n}^{4} - T_{e n v i r o n m e n t}^{4})

Let,

σ e (T_{p e n g u i n}^{4} - T_{e n v i r o n m e n t}^{4}) = k

∴ H e a t l o s t b y 1000 p e n g u i n s r a d i a t i n g f a r a w a y f r o m e a c h o t h e r, Q_{1}

= 1000 (2 π r_{p e n g u i n} h + 2 π r_{p e n g u i n}^{2}) k

Now,

π r_{p e n g u i n}^{2} = 0.26 m^{2} o r r_{p e n g u i n} = 0.287 m

Also,

π r_{h u d d l e}^{2} = N a_{p e n g u i n} = 1000 \times 0.26 o r r_{h u d d l e} = 9.09 m

∴ H e a t

lost by hudle of [enguins,

Q_{2}

= σ e A_{h u d d l e} (T_{p e n g u i n}^{4} - T_{e n v i r o n m e n t}^{4}) = (2 p i r_{h u d d l e} h + 2 π r_{h u d d l e}^{2}) k

NOw, percentage of reduce in radiation loss is given by

\frac{Q_{2}}{Q_{1}} = (2 p i r_{h u d d l e} h + 2 π r_{h u d d l e}^{2}) \times 100 /

N (2 p i r_{p e n g u i n} h + 2 π r_{h u d d l e}^{2}) = 26.67 %

Question 10. Variation of radiant energy emitted by sun, filament of tungsten lamp and welding arc as a function of its wavelength is shown in figure. Which of the following option is the correct match? (IIT JEE 2005)