We would like to make a vessel whose volume does not change with temperature. We can use brass and iron ( γ b r a s s = 6 × 10 − 5 K − 1 and γ i r o n = 3 .55 × 10 − 5 K − 1 ) to create a volume of 100 cc . How do you think you can achieve this?

Given: γ b r a s s = 6 × 10 − 5 K − 1 and γ i r o n = 3 .55 × 10 − 5 K − 1 We have to make double container whose difference in volume is 100 c c . Let the difference in initial volumes of containers of iron and brass be 100 cc . Hence: V I - V B = 100 cm 3 ⇒ V I - V B = 100 × 10 - 6 m 3 = 10 - 4 m 3 ⇒ V I - V B = 10 - 4 m 3 ....(i) Now as the containers are heated by Δ T K , and their new volumes become V I ' and V B ' , but their difference remains 100 cc. This condition is possible only if ∆ V I = ∆ V B .....(ii) The expansion in the volumes of the two containers will be: Δ V I = V I γ I Δ T I ....(iii) Δ V B = V B γ B Δ T B .....(iv) Also, ∆ T I = ∆ T B .....(v) On dividing (iii) and (iv) we get: ⇒ Δ V I ∆ V B = V I γ I Δ T I V B γ B Δ T B Using (iii) and(v) we get: ⇒ 1 = V I γ I V B γ B ⇒ V I γ I = V B γ B ⇒ V I = V B γ B γ I Substituting the given values of γ B and γ I ⇒ V I = V B 6 × 10 - 5 3 . 55 × 10 - 5 ⇒ V I = 6 3 . 55 V B ⇒ V I = 1 . 69 V B .....(vi) Substituting the value of V I in (i) we get: 1 . 69 V B - V B = 10 - 4 ⇒ 0 . 69 V B = 10 - 4 ⇒ V B = 10 - 4 0 . 69 = 144 . 9 × 10 - 6 m 3 ⇒ V B = 144 . 9 cc Substituting the above obtained value of V B in (vi): V I = 1 . 69 144 . 9 ⇒ V I = 244 . 9 cc Hence, we can make a double wall vessel whose volume does not change with temperature by taking an iron vessel of volume V I = 244 . 9 cc and by taking a brass vessel of volume V B = 144 . 9 cc.

MEDIUM

11th CBSE

IMPORTANT

Earn 100

Calculate the temperature which has same numeral value on celsius and Fahrenheit scale.

Important Questions on Thermal Properties of Matter

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 11>Chapter 11 - Thermal Properties of Matter>VSA>Q 11.17

These days people use steel utensils with copper bottom. This is supposed to be good for uniform heating of food. Explain this effect using the fact that copper is the better conductor.

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 11>Chapter 11 - Thermal Properties of Matter>SA>Q 11.18

Find out the increase in moment of inertia $(I)$ of a uniform rod (coefficient of linear expansion $α$ ) about its perpendicular bisector when its temperature is slightly increased by $Δ T$ .

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 11>Chapter 11 - Thermal Properties of Matter>SA>Q 11.19

During summers in India, one of the common practice to keep cool is to make ice balls of crushed ice, dip it in a flavoured sugar syrup and sip it. For this a stick is inserted into the crushed ice and is squeezed in the palm to make it into the ball. Equivalently in winter, in those areas wherever there is snow-fall, people make snow balls and throw around. Explain the formation of ball out of crushed ice or snow in the light of

P - T

diagram of water.

EASY

11th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 11>Chapter 11 - Thermal Properties of Matter>SA>Q 11.20

$100 g$ of water is supercooled to $- 10^{o} C$ . At this point, due to some disturbance mechanized or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze?

[ $s_{w} = 1 cal / g /^{0} C$ and $L_{F u s i o n} = 80 cal / g$ ]

EASY

11th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 11>Chapter 11 - Thermal Properties of Matter>SA>Q 11.21

One day in the morning, Ramesh filled up

1 / 3

bucket of hot water from geyser, to take bath. Remaining

2 / 3

was to be filled by cold water (at room temperature) to bring mixture to a comfortable temperature. Suddenly Ramesh had to attend to something which would take some times, say

5 - 10

minutes before he could take bath. Now he had two options:

(i)

fill the remaining bucket completely by cold water and then attend to the work,

(i i)

first attend to the work and fill the remaining bucket just before taking bath. Which option do you think would have kept water warmer? Explain.

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 11>Chapter 11 - Thermal Properties of Matter>LA>Q 11.22

We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say

10 cm .

We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their lengths remain constant. If

α_{i r o n} = 1.2 \times 10^{- 5} / K

and

α_{b r a s s} = 1.8 \times 10^{- 5} / K

, what should we take as length of each strip?

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 11>Chapter 11 - Thermal Properties of Matter>LA>Q 11.23

We would like to make a vessel whose volume does not change with temperature. We can use brass and iron ( $γ_{b r a s s} = 6 \times 10^{- 5} K^{- 1}$ and $γ_{i r o n} = 3 .55 \times 10^{- 5} K^{- 1}$ ) to create a volume of $100 cc .$ How do you think you can achieve this?

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Physics - Class 11>Chapter 11 - Thermal Properties of Matter>LA>Q 11.24

Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of

57 ° C

is drunk. You can take body (tooth) temperature to be

37 °

C and

α = 1.7 \times 10^{- 5} / K .

Bulk modulus for copper

= 140 \times 10^{9} N / m^{2}

Calculate the temperature which has same numeral value on celsius and Fahrenheit scale.

Important Questions on Thermal Properties of Matter

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