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

12th Grade > Physics

HEAT TRANSFER MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21. A body initially at

80^{\circ}

C cools to

64^{\circ}

C in 5 minutes and to

52^{\circ}

C in 10 minutes. The temperature of the body after 15 minutes will be

42.7∘ C
35∘ C
47∘ C
40∘ C

Discuss Question

Answer: Option A. -> 42.7∘ C
:
A
According to Newton law of cooling

\frac{θ_{1} - θ_{2}}{t} = K [\frac{θ_{1} - θ_{2}}{2} - θ_{0}]

A Body Initially At 80∘ C Cools To 64∘ C In 5 Minutes An...

For first process :

\frac{(80 - 64)}{5} = K [\frac{80 + 64}{2} - θ_{0}]

....(i)
For second process :

\frac{(80 - 52)}{10} = K [\frac{80 + 52}{2} - θ_{0}]

....(ii)
For third process :

\frac{(80 - θ)}{15} = K [\frac{80 + θ}{2} - θ_{0}]

....(iii)
On solving equation (i) and (ii) we get

K = \frac{1}{15}

and

θ_{0} = 24^{\circ} C

. Putting these values in equation (iii) we get

θ = {42.7}^{\circ} C

Question 22. The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are

T_{2} a n d T_{1} (T_{2} > T_{1})

. The rate of heat transfer through the slab, in a steady state is

(\frac{A (T_{2} - T_{1}) K}{x}) f

, with f which equal to

The Temperature Of The Two Outer Surfaces Of A Composite Sla...

Discuss Question

Answer: Option D. -> 13
:
D
Equation of thermal conductivity of the given combination

K_{e q} = \frac{l_{1} + l_{2}}{\frac{l_{1}}{K_{1}} + \frac{l_{2}}{K_{2}}} = \frac{x + 4 x}{\frac{x}{K} + \frac{4 x}{2 K}} = \frac{5}{3} K

. Hence rate of
flow of heat through the given combination is

\frac{Q}{t} = \frac{K_{e q} . A (T_{2} - T_{1}))}{(x + 4 x)} = \frac{\frac{5}{3} K A (T_{2} - T_{1})}{5 x} = \frac{\frac{1}{3} K A (T_{2} - T_{1})}{x}

On comparing it with given equation we get

f = \frac{1}{3}

Question 23. Two conducting rods A and B of same length and cross-sectional area are connected (i) In series (ii) In parallel as shown. In both combination a temperature difference of

100^{\circ}

C is maintained. If thermal conductivity of A is 3K and that of B is K then the ratio of heat current flowing in parallel combination to that flowing in series combination is

Two Conducting Rods A And B Of Same Length And Cross-section...

Discuss Question

Answer: Option A. -> 163
:
A
Heat current

H = \frac{Δ θ}{R} \Rightarrow \frac{H_{p}}{H_{s}} = \frac{R_{s}}{R_{p}}

In first case :

R_{s} = R_{1} + R_{2} = \frac{l}{(3 K) A} + \frac{l}{K A} = \frac{4}{3} \frac{l}{K A}

In second case :

R_{p} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{\frac{1}{(3 K) A} \times \frac{l}{K A}}{(\frac{l}{(3 K) A} + \frac{l}{K A})} = \frac{l}{4 K A}

∴ \frac{H_{p}}{H_{s}} = \frac{\frac{4 l}{3 K A}}{\frac{l}{4 K A}} = \frac{16}{3}

Question 24. Four identical rods of same material are joined end to end to form a square. If the temperature difference between the ends of a diagonal is

100^{\circ}

C , then the temperature difference between the ends of other diagonal will be

0∘C
100l∘C; where l is the length of each rod
1002l∘C
100∘C

Discuss Question

Answer: Option A. -> 0∘C
:
A
Suppose temperature difference between A and B is

100^{\circ} C

and

θ_{A} > θ_{B}

Four Identical Rods Of Same Material Are Joined End To End T...

Heat current will flow from A to B via path ACB and ADB. Since all the rod are identical so

Δ θ)_{A C} = (Δ θ)_{A D}

(Because heat current

H = \frac{Δ θ}{R}

; here R = same for all.)

\Rightarrow θ_{A} - θ_{C} = θ_{A} - θ_{D} \Rightarrow θ_{C} = θ_{D}

i.e. temperature difference between C and D will be zero.

Question 25. A solid copper sphere (density

ρ

and specific heat capacity (c) of radius r at an initial temperature 200K is suspended inside a chamber whose walls are at almost 0K. The time required (in

μ

s) for the temperature of the sphere to drop to 100 K is

807rρcσ
780rρcσ
277rρcσ
727rρcσ

Discuss Question

Answer: Option B. -> 780rρcσ
:
B

\frac{d T}{d t} = \frac{σ A}{m c J} (T^{4} - T_{0}^{4})

[In the given problem fall in temperature of body dT=(200-100)=100K , temp. of surrounding

T_{0} = 0 K

, Initial temperature of body T=200k].

\frac{100}{d t} = \frac{σ 4 π r^{2}}{\frac{4}{3} π r^{3} ρ c J} (200^{4} - 0^{4})

\Rightarrow d t = \frac{r ρ c J}{48 σ} \times 10^{- 6} s = \frac{r ρ c}{σ} . \frac{4.2}{48} \times 10^{- 6}

[As J = 4.2]

\frac{7}{80} \frac{r ρ c}{σ} μ s

Question 26. An ice box used for keeping eatable cold has a total wall area of

1 m e t r e^{2}

and a wall thickness of 5.0 cm. The thermal conductivity of the ice box is

K = 0.01 j o u l e / m e t r e -^{\circ} C

. It is filled with ice at

0^{\circ}

along with eatables on a day when the temperature is

30^{\circ}

. The latent heat of fusion of ice is

334 \times 10^{3} j o u l e s / k g

. The amount of ice melted in one day is
(1 day =86,400 seconds )

776 gms
7760 gms
11520 gms
1552 gms

Discuss Question

Answer: Option D. -> 1552 gms
:
D

\frac{d Q}{d t} = \frac{K A}{l} d θ = \frac{0.01 \times 1}{0.05} \times 30 = 6 J / s e c

Heat transferred in on day (86400 sec)

Q = 6 \times 86400 = 518400 J

Now

Q = m L \Rightarrow m = \frac{Q}{L} = \frac{518400}{334 \times 10^{3}}

= 1.552 k g = 1552 g

Question 27. The following figure shows two air-filled bulbs connected by a U-tube partly filled with alcohol. What happens to the levels of alcohol in the limbs X and Y when an electric bulb placed midway between the bulbs is lighted.

The level of alcohol in limb X falls while that in limb Y rises
The level of alcohol in limb X rises while that in limb Y falls
The level of alcohol falls in both limbs
There is no change in the levels of alcohol in the two limbs

Discuss Question

Answer: Option A. -> The level of alcohol in limb X falls while that in limb Y rises
:
A
Black bulb absorbs more heat in comparison with painted bulb. So air in black bulb expands more. Hence the level of alcohol in limb X falls while that in limb Y rises.

Question 28. One end of a copper rod of uniform cross-section and of length 3.1 m is kept in contact with ice and the other end with water at

100^{\circ}

C. At what point along it's length should a temperature of

200^{\circ}

C be maintained so that in steady state, the mass of ice melting be equal to that of the steam produced in the same interval of time. Assume that the whole system is insulated from the surroundings. Latent heat of fusion of ice and vaporisation of water are 80 cal/gm and 540 cal/gm respectively

21 cm from 100∘C end
21 cm from 0∘C end
125 cm from 100∘C end
125 cm from 0∘C end

Discuss Question

Answer: Option A. -> 21 cm from 100∘C end
:
A
Rate of flow of heat is given by

\frac{d Q}{d t} = \frac{Δ θ}{l / K A} a l s o \frac{d Q}{d t} = L \frac{d m}{d t}

(where L = Latent heat)

\Rightarrow \frac{d m}{d t} = \frac{K A}{l} (\frac{Δ θ}{L})

. Let the desire point is at a distance xfrom water at

100^{\circ}

One End Of A Copper Rod Of Uniform Cross-section And Of Leng...

∵

Rate of ice melting = Rate at which steam is being produced

\Rightarrow {(\frac{d m}{d t})}_{S t e a m} = {(\frac{d m}{d t})}_{I c e} \Rightarrow {(\frac{Δ θ}{L l})}_{S t e a m} = {(\frac{Δ θ}{L l})}_{I c e}

\Rightarrow \frac{(200 - 100)}{540 \times x} = \frac{(200 - 0)}{80 (3.1 - x)} \Rightarrow x = 0.21 m = 21 c m

Question 29. Four rods of identical cross-sectional area and made from the same metal form the sides of square. The temperature of two diagonally opposite points is T and