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Thermal Properties of Matter NEET Questions — Free MCQ Practice

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One kilogram of ice at $0 °$ C is mixed with one kilogram of water at 80 $° C$ . The final temperature of the mixture is (Take: Specific heat of water = 4200 J $k g^{- 1} k^{- 1}$ , Latent heat of ice = 336 kJ $k g^{- 1}$ )

A.

$0 °$ C

B.

$50 ° C$

C.

$40 ° C$

D.

$60 ° C$

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The fastest mode of heat transfer is

A.

Conduction

B.

Radiation

C.

Convection

D.

all are equally fast

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Explanation

In radiation, heat is transferred by electromagnetic waves, which have the fastest speed, so it is the fastest mode of heat transmission.

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$A temperature of 100 ° F (Farenhiet scale) is equal to T K (Kelvin scale) . The value of T is$

A.

310.9

B.

37.8

C.

100

D.

122.4

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A black body at temperature 300K radiates heat at the rate E. If its temperature is increased by 600K, the rate of radiation will increase to -

A.

16E

B.

64E

C.

81E

D.

256E

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Explanation

According to Stefan-Boltzmann law, the rate of radiation emitted by a black body is proportional to the fourth power of its absolute temperature. If the temperature increases from 300K to 900K, the rate of radiation increases by (900/300)^4 = 81 times.

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$A body cools down from 80 ° C to 70 ° C in 5 minutes . the temperature of the body will fall from 70 ° C to 60 ° C in a time$

A.

Less than 5 minutes

B.

equal to 5 minutes

C.

More than 5 minutes

D.

Can't say anything like the temperature of surrounding is not known

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Explanation

$T_{s} = T A p p l y i n g N e w t o n' s l a w o f c o o l i n g, F o r 1^{s t} c a s e - \frac{10}{5} = - b A (75 - T) . . . . (1) F o r 2^{n d} c a s e - \frac{10}{t} = - b A (65 - T) . . . . (2) f r o m (1) a n d (2) w e c a n s a y t h a t t > 5$

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Heat capacity is equal to the product of:

A.

mass and gas constant

B.

mass and specific heat

C.

latent heat and volume of water

D.

mass and Avogadro number

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Explanation

Heat capacity = mass x specific heat

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When a block of iron floats in Hg at $0 ° C$ , a fraction $K_{1}$ of its volumen= is submerged, while at temperature of $60 ° C$ a fraction $K_{2}$ is seen to be submersed. If the coefficient of volume expansion of iron is $γ_{F e}$ and that of mercury is $γ_{H g}$ , then the ratio $\frac{K_{1}}{K_{2}}$ can be expressed as:

A.

$\frac{1 + 60 γ_{F e}}{1 + 60 γ_{H g}}$

B.

$\frac{1 - 60 γ_{F e}}{1 + 60 γ_{H g}}$

C.

$\frac{1 + 60 γ_{F e}}{1 - 60 γ_{H g}}$

D.

$\frac{1 + 60 γ_{H e}}{1 - 60 γ_{F e}}$

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Explanation

$L e t V_{0} i s t h e v o l u m e o f t h e b l o c k . T h e n v o l u m e o f t h e b l o c k a t 60^{0} C - V = V_{0} (1 + γ_{i r o n} T) L e t V_{0}' i s t h e v o l u m e o f b l o c k s u b m e r g e d i n t h e m e r c u r y, t h e n v o l u m e o f t h e b l o c k s u b m e r g e d i n t h e m e r c u r y a t 60^{0} C - V' = V_{0}' (1 + γ_{m e r c u r y} T) (V o l u m e o f t h e b l o c k s u b m e r g e d i n m e r c u r y = v o l u m e o f m e r c u r y d i s p l a c e d) T h e n, \frac{V}{V'} = \frac{V_{0} (1 + γ_{i} T)}{V_{0}' (1 + γ_{m} T)} \Rightarrow \frac{1}{K_{2}} = \frac{1}{K_{1}} \frac{(1 + γ_{i} T)}{(1 + γ_{m} T)} \Rightarrow \frac{K_{1}}{K_{2}} = \frac{(1 + γ_{i} T)}{(1 + γ_{m} T)} = \frac{(1 + 60 γ_{i})}{(1 + 60 γ_{m})}$

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Hot water cools from 60 $° C$ to 50 $° C$ in first 10 minutes and from 50 $° C$ to 42 $° C$ in next 10 minutes. The temperature of surrounding is :

A.

$5 ° C$

B.

$10 ° C$

C.

$15 ° C$

D.

$20 ° C$

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Explanation

$(2) (\frac{θ_{1} - θ_{2}}{t}) = C (\frac{θ_{1} + θ_{2}}{2} - θ_{0}) \frac{10}{10} = C [55 - θ_{0}] (1) \frac{8}{10} = C [46 - θ_{0}] (2) \frac{10}{8} = \frac{55 - θ_{0}}{46 - θ_{0}} 460 - 10 θ_{0} = 440 - 8 θ_{0} 20 = 2 θ_{0} θ_{0} = 10 ° C$

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$If α_{c} and α_{f} denote the numerical values of coefficient of linear expansion of a solid, expressed per ° C and per ° F respectively, then$

A.

$α_{c} > α_{f}$

B.

$α_{f} > α_{c}$

C.

$α_{f} = α_{c}$

D.

$α_{f} + α_{c} = 0$

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Explanation

$(1) α = \frac{∆ L}{L ∆ T} α \propto \frac{1}{∆ T}$

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$A pendulum clock runs fast by 5 \sec . per day at 20 ° C and goes slow by 10 \sec . per day at 35 ° C . It shows correct time at a temperature of$

A.

$27.5 ° C$

B.

$25 ° C$

C.

$30 ° C$

D.

$33 ° C$

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Explanation

$(2) \frac{1}{2} α (35 - T) \times 86400 = 10 . . . (1) \frac{1}{2} α (T - 20) \times 86400 = 5 . . . (2) Solving (1) and (2) T = 25^{0} C .$

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

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