CLASS : XITH SUBJECT : PHYSICS
DATE : Solutions DPP NO. : 9
Topic :- THERMAL PROPERTIES OF MATTER
1 (a)
When a piece of glass is heated, due to low thermal conductivity it does not conduct heat
fast. Hence unequal expansion of it’s layers crack the glass
2 (a)
Latent heat is independent of configuration. Ordered energy spent in stretching the spring
will not contribute to heat which is disordered kinetic energy of molecules of substance
3 (c)
𝑇1 𝜆𝑚2 5.5 × 105 1 1
= = 5
= ⇒ 𝑛 = [Given 𝑇1 = 𝑛𝑇2 ]
𝑇2 𝜆𝑚1 11 × 10 2 2
4 (c)
Ice (0℃) converts into steam (100℃) in following three steps.
Total heat required 𝑄 = 𝑄1 + 𝑄2 + 𝑄3
= 5 × 80 + 5 × 1 × (100 − 0) + 5 × 540 = 3600 𝑐𝑎𝑙
Ice
0°C (Q1 = mLi)
Water at 0°C
(Q2 = mcW)
(Q3 = mLV)
Steam at 100°C Water at 100°C
6 (b)
𝜃1 −𝜃2 𝜃1 +𝜃2
According to Newton’s law = 𝑘[ − 𝜃0 ]
𝑡 2
Initially,
(80 − 64) 80 + 64
= 𝐾( − 𝜃0 ) ⇒ 3.2 = 𝐾(72 − 𝜃0 ) . . . (i)
5 2
Finally
(64 − 52) 64 + 52
= 𝐾[ − 𝜃0 ] ⇒ 1.2 = 𝐾[58 − 𝜃0 ] . . . (ii)
10 2
On solving equation (i) and (ii), 𝜃0 = 49℃
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7 (a)
Let the common temperature is 𝑥 on both scales.
𝐶 𝐹−32
5
= 9
Put 𝐶 = 𝐹 = 𝑥
𝑥 𝑥 − 32
∴ =
5 9
Or 9𝑥 = 5𝑥 − 160
Or 4𝑥 = −160
∴ 𝑥 = −40℃
8 (c)
𝑑𝑄 𝑑𝜃
= −𝐾𝐴
𝑑𝑡 𝑑𝑥
𝑑𝑄
∵ 𝑑𝑡 , 𝐾 and 𝐴 are constants for all points
⇒ 𝑑𝜃 ∝ −𝑑𝑥; 𝑖. 𝑒., temperature will decrease linearly with 𝑥
9 (a)
The contraction in the length of the wire due to change in
temperature=α𝐿𝑇 = 1.2 × 10−5 × 3 × (−170 − 30)
= −7.2 × 10−3 m
The expansion in the length of wire due to stretching force
𝐹𝐿 (10×10)×3
= 𝐴𝑌 = (0.75×10−6 )(2×1011 )
−3
= 2 × 10 m
Resultant change in length
= −7.2 × 10−3 + 2 × 10−3
= −5.2 × 10−3m= −5.2mm
Negative sign shows a contradiction.
11 (a)
𝐾1 𝑙12 10 2 1
𝐾 ∝ 𝑙2 ⇒ = 2=( ) =
𝐾2 𝑙2 25 6.25
12 (a)
Hear received by end 𝐴, for melting of ice
𝐾 𝐴(400−0)𝑡
𝑄𝐴 = = 𝑚 𝐿𝑖𝑐𝑒 …(i)
𝜆∙𝑥
Heat received by end 𝐵, for vaporization of water
𝐾 𝐴(400−100)𝑡
𝑄𝐵 = = 𝑚 𝐿𝑣𝑎𝑝 …(i)
(10−𝜆)𝑥
400
𝐿𝑖𝑐𝑒
Dividing both equation, 𝜆∙𝑥
300 =
𝐿𝑣𝑎𝑝
(10−𝜆)𝑥
4 (10 − 𝜆) 80
⇒ = ⇒𝜆=9
3 𝜆 540
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13 (a)
Freezing point of water decreases when pressure increases, because water expands on
solidification. “Except water” for other liquid freezing point increases with increase in
pressure.
Since the liquid in question is water. Hence, it expands on freezing
14 (a)
Thermal conductivity is independent of temperatures of the wall, it is a constant for the
material, so it will remain unchanged
15 (d)
𝛾real = 𝛾app + 𝛾vessel ; 𝛾vessel = 3𝛼
For vessel ′𝐴′ ⇒ 𝛾real = 𝛾1 + 3𝛼
For vessel ′𝐵′ ⇒ 𝛾real = 𝛾2 + 3𝛼𝐵
𝛾1 − 𝛾2
Hence, 𝛾1 + 3𝛼 = 𝛾2 + 3𝛼𝐵 ⇒ 𝛼𝐵 = +𝛼
3
16 (c)
𝑑𝜃 𝜀𝐴𝜎 3
= 4𝜃 ∆𝜃
𝑑𝑡 𝑚𝑐 0
𝜀𝐴𝜎
For given sphere and cube 𝑚𝑐 4𝜃03 ∆𝜃 is constant so for both rate of fall of temperature
𝑑𝜃
𝑑𝑡
= constant
17 (b)
∆𝑇 1 1
Loss in time per second 𝑇
= 2 𝛼∆𝜃 = 2 𝛼(𝑡 − 0)
⇒ loss in time per day
1 1 1
∆𝑡 = ( 𝛼𝑡) 𝑡 = 𝛼𝑡 × (24 × 60 × 60) = 𝛼𝑡 × 86400
2 2 2
18 (a)
𝐶𝑢 is better conductor than 𝐴𝑙 and 𝐴𝑔 is better conductor than 𝐶𝑢. Hence conductivity in
increasing order is 𝐴𝑙 < 𝐶𝑢 < 𝐴𝑔
19 (a)
𝐾1 𝜃1 +𝐾2 𝜃2
Temperature of interface 𝜃 = 𝐾1 +𝐾2
𝐾 1
[∵ 𝐾1 = ⇒ If 𝐾1 = 𝐾 then 𝐾2 = 4𝐾]
2 4
𝐾 × 0 + 4𝐾 × 100
⇒𝜃= = 80℃
5𝐾
20 (a)
Change in volume, ∆𝑉 = 𝑉𝛾 ∆𝑡
⇒ 0.24 = 100 × 𝛾 × 40
0.24
𝛾 = 100×40
= 0.00006 = 6 × 10−5
𝛾
𝛼= 3
⇒ 𝛼 = 2 × 10−5 ℃−1
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ANSWER-KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. A A C C A B A C A D
Q. 11 12 13 14 15 16 17 18 19 20
A. A A A A D C B A A A
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