IISER Physics - Thermal Properties of Matter
Exam Duration: 45 Mins Total Questions : 20
Heat given to a system can be associated with
- (a)
kinetic energy of random motion of molecules
- (b)
kinetic energy of orderly motion of molecules
- (c)
total kinetic energy of random and orderly motion of molecules
- (d)
kinetic energy of random motion in some cases and kinetic energy of orderly motion in other
Heat energy given to a body is due to kinetic energy of random motion of molecules.
At what temperature (in o C), the fahrenheit and celsius scale gives same reading?
- (a)
40
- (b)
-40
- (c)
8
- (d)
-8
We know that,
\(\frac { { T }_{ c }-0 }{ 100 } =\frac { { T }_{ F }-32 }{ 180 } \Rightarrow { T }_{ c }=\frac { 5 }{ 9 } \left( { T }_{ F }-32 \right) \\ Let,\quad \quad { T }_{ c }={ T }_{ F }=x\Rightarrow 9x=5x-32\times 5\\ \Rightarrow \quad \quad \quad x=\frac { -32\times 5 }{ 4 } =-40\)
-40o C and -40o F represent same temperature.
A bimetallic strip is made of aluminium and steel \(\left( { \alpha }_{ Al }>{ \alpha }_{ steel } \right) \) on heating, the strip will
- (a)
remain straight
- (b)
get twisted
- (c)
get twisted
- (d)
bend with aluminium on concave side
As, \({ \alpha }_{ Al }>{ \alpha }_{ steel }\)
In bimetallic strip on heating,
Aluminium strip will expand, more than that of steel strip.
So, aluminium strip will bend more on convex side and steel on concave side.
If a piece of metal is heated to temperature \(\theta \) and then allowed to cool in a room which is at temperature \(\theta _0\), the graph between the temperature T of the metal and time t will be closed to
- (a)
- (b)
- (c)
- (d)
According to Newton's cooling law, option (c) is correct.
The radius of metal sphere at room temperature T is R and the coefficient of linear expansion of the metal is \(\alpha \) . The sphere is heated a little by a temperature T so that, new temperature is \(T+\Delta T.\) The increase in volume of sphere is approximately
- (a)
\(2\pi R\alpha \Delta T\)
- (b)
\(\pi { R }^{ 2 }\alpha \Delta T\)
- (c)
\(4\pi { R }^{ 3 }\alpha \Delta T/3\)
- (d)
\(4\pi { R }^{ 3 }\alpha \Delta T\)
Since, coefficient of volume expansion,
\(\gamma =\) 3x coefficient of linear expansion
\(\Rightarrow\) \(\gamma =\)3\(\alpha\)
Now, \(\frac { 1 }{ \Delta T } \left( \frac { \Delta V }{ V } \right) =3\alpha \)
\(\Rightarrow\) \(\Delta V=3V.\alpha .\Delta T\)
\(\Rightarrow\) \(\Delta V=\frac { 4 }{ 3 } \pi { R }^{ 3 }\times 3\alpha \times \Delta T=4\pi { R }^{ 3 }\alpha \Delta T\)
Which one of the following would raise the temperature of 40 g of water at 20o C most mixed with?
- (a)
20 g of water at 40o C
- (b)
30 g of water at 30o C
- (c)
10 g of water at 60o C
- (d)
4 g of water at 100o C
If m is the mass of water added at \(\theta\)oC, then resultant temperature T is given as,
Heat lost = Heat gained
\(\Rightarrow\) m(\(\theta\)-T) = 40 (T-20)
\(\Rightarrow\) T =\(\frac { 800+m.\theta }{ 40+m } \)
So, for option (a), T = \(\frac { 800+20\times 40 }{ 40+20 } ={ 26.6 }^{ o }C\)
For option (b), T = \(\frac { 800+30\times 30 }{ 40+30 } ={ 24.28 }^{ o }C\)
For option(c), T = \(\frac { 800+10\times 60 }{ 40+10 } ={ 28 }^{ o }C\)
For option (d), T=\(\frac { 800+400 }{ 40+4 } ={ 27.7 }^{ o }C\)
A black body is at 2000 K, it emits maximum energy at a wavelength 1\(\mu m\) . The temperature, it emits at wavelength of 2\(\mu m\) is
- (a)
1000 K
- (b)
2000 K
- (c)
500 K
- (d)
100 K
Here, T1 = 2000 K; \({ \lambda }_{ 1 }=1\mu m\)
T2=1;\({ \lambda }_{ 2 }=2\mu m\)
We know that, \({ \lambda }_{ 1 }{ T }_{ 1 }={ \lambda }_{ 2 }{ T }_{ 2 };2000\times 1={ T }_{ 2 }\times 2;{ T }_{ 2 }=1000K\)
A star emits wavelength 289.8 nm of maximum intensity. Then, radiant intensity of state is \((\sigma =5.67\times { 10 }^{ -8 }W/{ m }^{ 2 }/{ k }^{ 4 },\) Wein's constant, b=2898X10-6 mK)
- (a)
5.67X1016 W/m2
- (b)
5.67X1014 W/m2
- (c)
5.67X1010 W/m2
- (d)
5.67X108 W/m2
Wein's law,
\({ \lambda }_{ m }T=\quad b\quad constant\\ \Rightarrow T=\frac { b }{ { \lambda }_{ m } } =\frac { 2898\times { 10 }^{ -6 } }{ 2898\times { 10 }^{ -10 } } ={ 10 }^{ 4 }\\ Now,\quad u=\sigma { T }^{ 4 }\\ \Rightarrow u=5.67\times { 10 }^{ -8 }\times { 10 }^{ 16 }=5.67\times { 10 }^{ 8 }W/{ m }^{ 2 }\)
Wein's displacement law for emission of radiation can be written as
- (a)
\({ \lambda }_{ max }\) is inversely proportional to square of absolute temperature (T2)
- (b)
\({ \lambda }_{ max }\) is inversely proportional to absolute temperature (T)
- (c)
\({ \lambda }_{ max }\) is directly proportional to absolute temperature (T)
- (d)
\({ \lambda }_{ max }\) is inversely proportional to square of absolute temperature (T2 )
According to Wein's displacement law,
\({ \lambda }_{ max }.T=b\)
where, b=Wien's constant
\({ \lambda }_{ max }\propto \frac { 1 }{ T } \)
i.e., \({ \lambda }_{ max }\) is inversely proportional to absolute temperature.
If temperature of black body increases from 300 K to 900 K, then the rate of energy radiation increase by
- (a)
81
- (b)
3
- (c)
9
- (d)
2
We can write, \({ \left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }^{ 4 }=\frac { { E }_{ 2 } }{ { E }_{ 1 } } \)
or \(\frac { { E }_{ 2 } }{ { E }_{ 1 } } ={ \left( \frac { 900 }{ 300 } \right) }^{ 4 }\Rightarrow \frac { { E }_{ 2 } }{ { E }_{ 1 } } ={ \left( 3 \right) }^{ 4 }\)
or \({ E }_{ 2 }=81{ E }_{ 1 }\quad \Rightarrow \quad \frac { { E }_{ 2 } }{ { E }_{ 1 } } =81\)
A black body radiate energy at rate of X W/m2 at a high temperature of TK. When temperature is reduced to \(\left( \frac { T }{ 2 } \right) K\) , the radiant energy is
- (a)
\(\frac { X }{ 16 } \)
- (b)
\(\frac { X }{ 4 } \)
- (c)
\(\frac { X }{ 2 } \)
- (d)
\(2X\)
According to Stefan's law,
\(X\propto { T }^{ 4 }\quad \) (\(\because\) X2: Energy at T/2 K)
\(\Rightarrow\) \(\frac { { \quad X }_{ 2 } }{ { \quad X }_{ 1 } } ={ \left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }^{ 4 }\) (\(\because\) X1: Energy at T k)
or \(\frac { { \quad X }_{ 2 } }{ { \quad X }_{ 1 } } =\frac { { \left( \frac { T }{ 2 } \right) }^{ 2 } }{ { T }^{ 4 } } ={ \left( \frac { 1 }{ 4 } \right) }^{ 4 };{ \quad X }_{ 2 }=\frac { { \quad X }_{ 1 } }{ 16 } \)
Radiant energy is \(\frac { { \quad X } }{ 16 } \)
A black body radiates heat energy at the rate of 3X106 W at 127oC. The temperature at which it would radiate heat energy at 243X106 W is
- (a)
1000 K
- (b)
1200 K
- (c)
1400 K
- (d)
1600 K
\(By\quad Stefan's\quad law,\quad U=\sigma A{ T }^{ 4 }\\ So,\quad \frac { { U }_{ 1 } }{ { U }_{ 2 } } ={ \left( \frac { { T }_{ 1 } }{ { T }_{ 2 } } \right) }^{ 4 }\Rightarrow \frac { 3\times { 10 }^{ 6 } }{ 243\times { 10 }^{ 6 } } ={ \left( \frac { { T }_{ 1 } }{ { T }_{ 2 } } \right) }^{ 4 }\)
T2=3X(273+127) = 3X400 = 1200 K
Newton's law of cooling hold's good only, if the temperature difference between the body and the surrounding is
- (a)
less than 40o C
- (b)
more than 50o C
- (c)
less than 100o C
- (d)
more than 100o C
Newton's law of cooling \(\left[ \frac { dT }{ dt } \propto \left( { \theta }_{ 1 }-{ \theta }_{ 0 } \right) \right] \) is valid for small temperature difference i.e., less than 40o C.
Hot water cools from 60o C to 50o C in first 10 min and 42o C in next 10 min. Then, the temperature of surrounding is
- (a)
10oC
- (b)
15oC
- (c)
20oC
- (d)
30oC
As per Newton,s law of cooling
\({ T }_{ 2 }-{ T }_{ 1 }=k\left[ \frac { { T }_{ 1 }+{ T }_{ 2 } }{ 2 } -{ T }_{ s } \right] \)
where Ts is the temperature of the surrounding.
\(\frac { 60-50 }{ 10 } =k\left[ \frac { 60+50 }{ 2 } -{ T }_{ s } \right] \)
\(\Rightarrow \quad \quad 1=k\left[ 55-{ T }_{ s } \right] \quad \quad \quad \quad ...(i)\)
Similarly, \(\frac { 50-42 }{ 10 } =k\left( 46-{ T }_{ s } \right) ;\frac { 8 }{ 10 } =k\left( 46-{ T }_{ s } \right) \quad ...(ii)\)
Dividing Eqs. (i) and (ii), we have
\(\frac { 10 }{ 8 } =\frac { k\left( 55-{ T }_{ s } \right) }{ k\left( 46-{ T }_{ s } \right) } \)
\(\Rightarrow\) \({ T }_{ s }={ 10 }^{ o }C\)
A clock with a steel pendulum keeps correct time at 25oC. What will be the error in second per day, if room temperature is 35oC? (coefficient of linear expansion, \({ \alpha }_{ steel }=5\times { 10 }^{ -5 }{ C }^{ -1 })\)
- (a)
43.2 s gained
- (b)
43.2 s lost
- (c)
21.6 s gained
- (d)
21.6 s lost
Change in temperature = 35 - 25 = 10o C
\(\therefore\) Change in length = \(\alpha \left( \Delta T \right) \times l\)
\(\Rightarrow\) Time lost = \(\frac { 1 }{ 2 } \times 5\times { 10 }^{ -5 }\times 10\times 86400\)
= \(21.6s\quad \left( \because \quad t\propto { l }^{ { -1 }/{ 2 } }\therefore \frac { \Delta t }{ t } =\frac { -1 }{ 2 } .\frac { \Delta l }{ l } \right) \)
If a liquid cools from 70oC to 55oC in 5 min and then to 45o C in 10 min, then temperature of surrounding is
- (a)
50oC
- (b)
60oC
- (c)
25oC
- (d)
20oC
From 70o C to 55o C
Change in temperature, \(\Delta\)T = 70-55 = 15oC
Time interval, \(\Delta\)t = 5min
Average temperature, T = \(\frac { 70+55 }{ 2 } ={ 62.5 }^{ o }C\)
Let, temperature of surrounding = Tsurr
\(\frac { \Delta T }{ \Delta t } =-K\left( T-{ T }_{ surr } \right) \)
\(\frac { 15 }{ 5 } =-k\left( 62.5-{ T }_{ surr } \right) \quad \quad \quad ...(i)\)
From \({ 55 }^{ o }C\) to \({ 45 }^{ o }C\) , \(\Delta T=55-45=10^{ o }C\)
\(\Rightarrow\) \(\Delta t=10-5=5min\)
Average temperature, \(T=\frac { 55+45 }{ 2 } =50^{ o }C\)
Now, \(\frac { 10 }{ 5 } =-k\left( 50-{ T }_{ surr } \right) \quad \quad ...(ii)\)
Dividing Eq.(i) by Eq.(ii) and solving, we get
\({ T }_{ surr }=25^{ o }C\)
The thickness of ice on a lake is 10 cm and the temperature of air is -10oC. If rate of cooling of water inside lake is 20000 cal min-1 through each square metre surface, then K for ice is
- (a)
14
- (b)
10
- (c)
3
- (d)
4
Given, thickness of ice, \(\Delta\)x = 10cm = 0.1m;
\(\Delta\)T=0-(-10)=10oC
\(\frac { \Delta Q }{ \Delta t } =20000cal{ \quad min }^{ -1 }=\frac { 20000 }{ 60 } cal{ \quad s }^{ -1 }\)
\(=\frac { 20000 }{ 60 } \times 4.2=1399.9\approx 1400{ Js }^{ -1 }\)
A=1m2, K=?
Now, \(\frac { \Delta Q }{ \Delta t } =KA\left( \frac { \Delta T }{ \Delta x } \right) \)
\(\therefore\) \(K=\frac { \frac { \Delta Q }{ \Delta t } }{ A\left( \frac { \Delta T }{ \Delta x } \right) } =\frac { 1400 }{ 1\left( \frac { 10 }{ 0.1 } \right) } =14W\)oC-1
If a cylinder of radius R having thermal conductivity K1 is surrounded by other cylindrical shell of radius 2R having thermal conductivity K2 . Two ends are maintained at two different temperatures. In steady state, the effective thermal conductivity of system is (assume no heat loss)
- (a)
\(\frac { { k }_{ 1 }+{ 4k }_{ 2 } }{ 4 } \)
- (b)
\(\frac { { k }_{ 1 }+{ 3k }_{ 2 } }{ 4 } \)
- (c)
\(\frac { { 4k }_{ 1 }+{ k }_{ 2 } }{ 4 } \)
- (d)
\(\frac { { 3k }_{ 1 }+{ k }_{ 2 } }{ 4 } \)
In steady state, \(\left( \frac { dQ }{ dt } \right) =\left( \frac { { dQ }_{ inner } }{ dt } \right) +\left( \frac { { dQ }_{ outer } }{ dt } \right) \)
\(\Rightarrow \frac { k\pi { \left( 2R \right) }^{ 2 }\left( { \theta }_{ 1 }-{ \theta }_{ 2 } \right) }{ l } =\frac { { k }_{ 1 }\pi { \left( R \right) }^{ 2 }\left( { \theta }_{ 1 }-{ \theta }_{ 2 } \right) }{ l } +{ k }_{ 2 }\frac { \left[ \pi { \left( 2R \right) }^{ 2 }-\pi { \left( R \right) }^{ 2 } \right] \left( { \theta }_{ 1 }-{ \theta }_{ 2 } \right) }{ l } \)
or \(\frac { k4\pi { \left( R \right) }^{ 2 }\left( { \theta }_{ 1 }-{ \theta }_{ 2 } \right) }{ l } =\frac { { k }_{ 1 }\pi { \left( R \right) }^{ 2 }\left( { \theta }_{ 1 }-{ \theta }_{ 2 } \right) }{ l } +\frac { { k }_{ 2 }3\pi { \left( R \right) }^{ 2 }\left( { \theta }_{ 1 }-{ \theta }_{ 2 } \right) }{ l } \)
or \(4k={ k }_{ 1 }+3{ k }_{ 2 }\)
or \(k=\frac { { k }_{ 1 }+3{ k }_{ 2 } }{ 4 } \)
Two rods X and Y having equal lengths. Then, cross-sectional area are Ax and Ay and thermal conductivities Kx and Ky . When the temperature at ends of each rod are are Tx and Ty respectively, the rate of flow of heat through X and Y will b, if equal
- (a)
\(\frac { { A }_{ x } }{ { A }_{ y } } =\frac { { K }_{ y } }{ { K }_{ x } } \)
- (b)
\(\frac { { A }_{ x } }{ { A }_{ y } } =\frac { { K }_{ y } }{ { K }_{ x } } \times \frac { { T }_{ y } }{ { T }_{ x } } \)
- (c)
\(\frac { { A }_{ x } }{ { A }_{ y } } =\sqrt { \frac { { K }_{ y } }{ { K }_{ x } } } \)
- (d)
\(\frac { { A }_{ x } }{ { A }_{ y } } ={ \left( \frac { { K }_{ y } }{ { K }_{ x } } \right) }^{ 2 }\)
\(We\quad can\quad write,\left( \frac { \Delta Q }{ \Delta t } \right) _{ x }=\left( \frac { \Delta Q }{ \Delta t } \right) _{ y }\\ \Rightarrow \frac { { k }_{ x }{ A }_{ x }\left( { T }_{ x }-{ T }_{ y } \right) }{ l } =\frac { { k }_{ y }{ A }_{ y }\left( { T }_{ x }-{ T }_{ y } \right) }{ l } \\ or\quad { k }_{ x }{ A }_{ x }={ k }_{ y }{ A }_{ y }\quad or\quad \frac { { A }_{ x } }{ { A }_{ y } } =\frac { { K }_{ y } }{ { K }_{ x } } \)
A liquid in a breaker has temperature \(\theta (t)\) at time t and \({ \theta }_{ 0 }\) is temperature of surroundings, then according to Newton's law of cooling the correct graph between \({ log }_{ e }\left( \theta -{ \theta }_{ 0 } \right) \) and t is
- (a)
- (b)
- (c)
- (d)
According to Newton's law of cooling, "rate of fall in temperature is proportional to the difference in temperature of the body with surrounding", i.e.,
\(\frac { -d\theta }{ dt } =h\left( \theta -{ \theta }_{ 0 } \right) \)
\(\Rightarrow \int { \frac { d\theta }{ \theta -{ \theta }_{ 0 } } } =\int { -kdt } \)
\(\Rightarrow ln\left( \theta -{ \theta }_{ 0 } \right) =kt+C\)
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