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Thermal Properties of Matter NEET Questions — Free MCQ Practice

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Certain quantity of water cools from 70°C to 60°C in the first 5 min and to 54°C in the next 5 min. The temperature of the surroundings is

A.

45°C

B.

20°C

C.

42°C

D.

10°C

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Explanation

Let the temperature of the surrounding is t°C

For first case, (70-60)/5min=K(65°-t°C)

(65° is average of 70°C and 60°C)

10/5min=K(65°C-t°C) ...(i)

For second case,

(60-54)/5min=K(57-t) ...(ii)

(57°C is average of 60°C and 54°C)

From Eqs. (i)/(ii), 10/6=(65-t)/(57-t)

Solving, we get t=45° C

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A piece of iron is heated in a flame. If first becomes dull red then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using

A.

Stefan's law

B.

Wien's displacement law

C.

Kirchoff's law

D.

Newton's law of cooling

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Explanation

(b) The equation of Wien's displacement law, i.e., λ mT=constant

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If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q?

A.

$Q / 4 {πR}^{2} σ$

B.

${(Q / 4 {πR}^{2} σ)}^{- 1 / 2}$

C.

${(4 {πR}^{2} Q / σ)}^{1 / 4}$

D.

${(Q / 4 {πR}^{2} σ)}^{1 / 4}$

$(σ s \tan d s f o r S t e f a n' s c o n s \tan t .)$

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Explanation

From Stefan' law,

$E = σ T^{4}$

So, the rate energy production

$Q = E \times A$

$Q = σ T^{4} \times 4 {πR}^{2}$

Temperature of star

$T = {(\frac{Q}{4 {πR}^{2} σ})}^{1 / 4}$

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A slab of stone of area of 0.36 $m^{2}$ and

thickness 0.1 m is exposed on the lower

surface to steam at $100 ° C$ . A block of ice

at $0 ° C$ rests on the upper surface of the slab.

In one hour 4.8 kg of ice is melted. The thermal

conductivity of slab is

(Given latent heat of fusion of ice

=3.36x $10^{5} J {kg}^{- 1}$ )

A.

$1.24 J / m / s / ° C$

B.

$1.29 J / m / s / ° C$

C.

$2.05 J / m / s / ° C$

D.

$1.02 J / m / s / ° C$

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Explanation

$\frac{δQ}{δt} = \frac{KA}{L} (T_{1} - T_{2}) Q = \frac{KA}{L} (T_{1} - T_{2}) t Q = {mL}_{f} \frac{KA}{L} (T_{1} - T_{2}) t = {mL}_{f} K = \frac{{mL}_{f} L}{A (T_{1} - T_{2}) t} K = \frac{4.8 \times 3.36 \times 10^{5} \times 0.1}{0.36 \times 100 \times 3600} = \frac{4.8 \times 3.36}{0.36 \times 36} = 1.24 J / m / s / ° C$

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When 1 kg of ice at $0 ° C$ melts to water at $0 ° C$ , the resulting change in its entropy, taking latent heat of ice to be $80 cal / ° C,$ is

A.

$8 \times 10^{4} cal / K$

B.

$80 cal / K$

C.

$293 cal / K$

D.

$273 cal / K$

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Explanation

Change in entropy

$∆ s = \frac{m L}{T} = \frac{1000 \times 80}{273} = 293 cal K^{- 1}$

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A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod when placed in thermal contact with the two reserviors in time t ?

A.

Q/4

B.

Q/16

C.

2Q

D.

Q/2

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Explanation

In steady state the amount of heat flowing from one face to the other face in time t is given by $Q = \frac{K A (θ_{1} - θ_{2}) t}{l}$ where K is coefficient of thermal conductivity of material of rod

$\Rightarrow \frac{Q}{t} \propto \frac{A}{l} \propto \frac{r^{2}}{l}$ ...(i)

As the metallic rod is melted and the material is formed into of half the radius

$V_{1} = V_{2}$

${πr}_{1}^{2} l_{1} = {πr}_{2}^{2} l_{2}$

$\Rightarrow l_{1} = \frac{l_{2}}{4}$ ...(ii)

Now, from Eqs. (i) and (ii)

$\frac{Q_{1}}{Q_{2}} = \frac{r_{1}^{2}}{l_{1}} \times \frac{l_{2}}{r_{2}^{2}} = \frac{r_{1}^{2}}{l_{1}} \times \frac{4 l_{1}}{{(r_{1} / 2)}^{2}}$

$\Rightarrow Q_{1} = 16 Q_{2}$

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The total radiant energy per unit area per unit time, normal to the direction of incidence, received at a distance R from the centre of a star of radius r,whose outer surface radiates as a black body at a temperature TK is given by

A.

$σ r^{2} T^{4} / R^{2}$

B.

$σ r^{2} T^{4} / 4 {πr}^{2}$

C.

$σ r^{4} T^{4} / r^{4}$

D.

$4 π σ r^{2} T^{4} / R^{2}$

(where $σ$ is Stefan's constant)

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Explanation

If r is the radius of the star and T its temperature, then the energy emitted by the star per second through radiation in accordance with Stefan's law will be given by

$P = A σ T^{4} = 4 {πr}^{2} {σT}^{4}$

In reaching a distance R this energy will spread over a sphere of radius R; so the intensity of radiation will be given by- $S = \frac{P}{A} = \frac{P}{4 {πR}^{2}} = \frac{4 {πr}^{2} {σT}^{4}}{4 {πR}^{2}}$

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A black body at $227 ° C$ radiates heat at the rate of 7 cal $c m^{- 2} s^{- 1} .$ At a temperature of $727 ° C,$ the rate of heat radiated in the same units will be

A.

60

B.

50

C.

112

D.

80

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Explanation

Apply Stefan's law of radiation.

$E = σ T^{4}$

$\frac{E_{1}}{E_{2}} = {[\frac{T_{1}}{T_{2}}]}^{4}$

$\Rightarrow E_{2} = 7 {[\frac{273 + 727}{273 + 227}]}^{4}$

= ${(\frac{1000}{500})}^{4} \times 7$

= $112 c a l ‐ c m^{- 2} s^{- 1}$

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The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures $T_{1} a n d T_{2} (T_{1} > T_{2}) .$ The rate f heat transfer, $\frac{d Q}{d t},$ through the rod in a steady state is given by

A.

$\frac{d Q}{d t} = \frac{K L (T_{1} - T_{2})}{A}$

B.

$\frac{d Q}{d t} = \frac{K L (T_{1} - T_{2})}{L A}$

C.

$\frac{d Q}{d t} = K L A (T_{1} - T_{2})$

D.

$\frac{d Q}{d t} = \frac{K A (T_{1} - T_{2})}{L}$

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Explanation

For a rod of length L and area of cross-section A whose faces are maintained at temperature $T_{1} a n d T_{2}$ respectively.

Then in steady state the rate of heat flowing from one face to the other face in time t is given by

$\frac{d Q}{d t} = \frac{K A (T_{1} - T_{2})}{L}$

The curved surface of rod is kept insulated from surrounding to avoid leakage of heat.

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

A.

$78 ° W$

B.

$117 ° W$

C.

$200 ° W$

D.

$139 ° W$

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Thermal Properties of Matter MCQs for NEET — Physics Questions with Answers

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