Q: The 'Y' of a material of a rod is 200 GPa. Its coefficient of linear expansion is 15 x 10 ^{-6}/^{0} C. It is fixed at two ends and is cooled by 100 ^{o} C. The thermal stress produced in the rod (in N.m ^{-2} )is :

Y = \frac{stress}{strain} stress = Y strain = Y \frac{\triangle l}{l} = Y \alpha \triangle T (\because \frac{\triangle l}{l} = \alpha \triangle T) = 200 \times 10^{9} \times 15 \times 10^{-6} \times 100 = 3 \times 10^{8} N/m^{2}

Q: A brass rod at 30^{\circ}C is observed to be 100 cm long when measured by a steel scale which is correct at 0^{\circ} C . \alpha for steel is 12\times 10^{-6}/^{\circ}C and \alpha for brass is 19\times 10^{-6}/^{\circ}C . The correct length of brass rod at 0^{\circ}C is :

As steel scale is correct at 0℃. Actual length measured by steel scale at 30℃. L=100(1+t\times \alpha )\\ L=100(1+12\times { 10 }^{ -6 }\times 30)\\ L=100(1.00036)cm\\ L=100.036cm Brass rod has length 100.036cm at 30℃ let's assume length L' is length at 0℃so - 100.036=L'(1+t\times \alpha )\\ 100.036=L'(1+19\times { 10 }^{ -6 }\times 30)\\ 100.036=L'(1.00057)\\ L'=99.979cm

Q: A glass bulb of volume 250cc is filled with mercury at 20 ^{0} C and the temperature is raised to 100^{0}C . If the coefficient of linear expansion of glass is 9\times 10^{-6/0}C Coefficient of absolute expansion of mercury is 18\times 10^{-4/0}C . The volume of mercury that overflows is

\gamma_{glass} =3\alpha_{glass}=3 \times 9 \times 10^{-6}=27 \times 10^{-6} \gamma_{mercury} = 18 \times 10^{-4} Amount of mercury to overflow from the glass V_{overflow} = V_0 \times( \gamma_{mercury} -\gamma_{glass} ) \times \Delta T= 250 \times (18 \times 10^{-4} - 27 \times 10^{-6}) \times (100-20)= 250 \times 1773 \times 10^{-6} \times 80 \implies V_{overflow}=35.46cc

Q: A vessel is half filled with a liquid at 0 ^{0} C. When then vessel is heated to 100 ^{0} C the liquid occupies 3/4 volume of the vessel. Coefficient of apparent expansion of the liquid is :

{\gamma}_a=\dfrac{V_2-V_1}{V_1 \times (t_2-t_1)} ....(1)Given, V_1=\dfrac{1}{2} V_2=\dfrac{3}{4} t_2-t_1=100-0=100 Putting the values in (1) {\gamma}_a= \dfrac{(\dfrac{3}{4}-\dfrac{1}{2})}{\dfrac {1}{2} \times 100}=0.005

Question 1

5\ g  of steam at  100^{0}C  is mixed with  5\ g  of the ice  at  0^{0}C . The final temperature of the mixture is :

Accepted Answer

Latent heat of fusion is  336\ J/g  Latent heat of vaporization is  2260\ J/g  So, Heat released by  5g  of steam to condense is  Q_1=5	imes10^{-3}	imes 2260	imes10^3 and, Heat absorbed by  5g  of ice to melt is  Q_2=5	imes10^{-3}	imes 336	imes10^3 Heat require to increase temperature from  0^oC\ to\ 100^oC  is  Q_3=mc\Delta T=5	imes10^{-3}	imes4200	imes100=2100\ J As,  Q_1>Q_2+Q_3 Hence, final temperature is at  100^oC .Partial steam will condense.

Question 2

Steam at  100^{\circ}C  is passed into  1.1  kg of water contained in a calorimeter of water equivalent  0.02  kg at  15^{\circ}C  till the temperature of the calorimeter and its contents rises to  80^{\circ}C .The mass of the steam condensed in kg is

Accepted Answer

Let  M  be the mass of steam condensed.Amount of heat lost by the steam due to latent heat of condensation and due to cooling to  80^{\circ}C =M	imes 540+m(100-80) Amount of heat absorbed by the water and calorimeter system= 1100	imes 1	imes (80-15)+20	imes 1 	imes (80-15) From conservation of heat energy,  =M	imes 540+M(100-80)   =1100	imes 1	imes (80-15)+20	imes 1 	imes (80-15) \implies M=130g=0.130kg

Question 3

10\ g  of Ice at - 20^{0}C  is added to  10g  of water at  50^{0}C . The amount of ice and water, present at equilibrium, are respectively

Accepted Answer

Specific heat of ice is  2.1\ J/g{}^oC Total heat released by water is  10	imes 4.2	imes50=2100\ J Total heat absorbed by ice from  -20^oC\ to\ 0^oC=10	imes2.1	imes20=420\ J \Delta Q=2100-420=mL Melted ice,  m=(2100-420)/336\approx5gm Hence, at equilibrium, total water is  15g , total ice is  5g .

Question 4

A pendulum clock shows correct time at  0^{0} C. At a higher temperature the clock

Accepted Answer

The time period pendulum is,  T = 2\pi\sqrt{(L/g)}  (where, T is the period , g=9.81 m/s/s) Since the length(L) of the pendulum changes with temperature so the period will also change. For higher temperature the length will increase and the period will also extend which shows the time of the ticks. So the clock will lose the time.

Question 5

The 'Y' of a material of a rod is 200 GPa. Its coefficient of linear expansion is 15 x 10 ^{-6}/^{0} C. It is fixed at two ends and is cooled by 100 ^{o} C. The thermal stress produced in the rod (in N.m ^{-2}  )is :

Accepted Answer

Y = \frac{stress}{strain}  stress = Y strain  = Y \frac{	riangle l}{l}  = Y  \alpha 	riangle T     (\because \frac{	riangle  l}{l} = \alpha 	riangle T)  = 200 	imes 10^{9} 	imes 15 	imes  10^{-6} 	imes 100  = 3 	imes 10^{8} N/m^{2}

Question 6

A brass rod at  30^{\circ}C  is observed to be 100 cm long when measured by a steel scale which is correct at  0^{\circ} C .  \alpha  for steel is  12	imes 10^{-6}/^{\circ}C  and  \alpha  for brass is  19	imes 10^{-6}/^{\circ}C . The correct length of brass rod at  0^{\circ}C  is :

Accepted Answer

As steel scale is correct at 0&#8451;. Actual length measured by steel scale at 30&#8451;. L=100(1+t	imes \alpha )\ L=100(1+12	imes { 10 }^{ -6 }	imes 30)\ L=100(1.00036)cm\ L=100.036cm  Brass rod has length 100.036cm at 30&#8451; let's assume length L' is length at 0&#8451;so - 100.036=L'(1+t	imes \alpha )\ 100.036=L'(1+19	imes { 10 }^{ -6 }	imes 30)\ 100.036=L'(1.00057)\ L'=99.979cm

Question 7

Two straight metallic strips each of thickness  t  and length  l  are riveted together. Their coefficients of linear expansion are  \alpha_{1}  and  \alpha_{2} . If they are heated through a temperature  	riangle t , the bimetallic strip will bend to form an arc of radius :

Accepted Answer

Given, l=  length of both the metallic stripes, t=  thickness of both the metallic stripes,let  l_1  and  l_2  be the final length of the metallic stripes afterexpansion due to  \Delta T  change. \Delta l=l_2-l_1=(l\alpha_2\Delta T-l\alpha_1\Delta T) We know, l=	heta r \Delta l=	heta \Delta r    Here,  	heta  is constant as same arc angle is made by both the stripes.and,  \Delta r=t  then  	heta=\dfrac{l(\alpha_2-\alpha_1)\Delta T}{t}=\dfrac{l}{r} r=\dfrac{t}{(\alpha_2-\alpha_1)\Delta T} Hence option  	extbf B  is the correct answer.

Question 8

Two rods P and Q of different metals having the same area A and the same length L are placed between two rigid walls as shown in figure. The coefficients of linear expansion of P and Q are  \alpha_{1}  and  \alpha_{2}   respectively and their Youngs modulii are  Y_{1}  and  Y_{2}  . The  temperature of both rods is now raised by T degrees. The new length of rod Q is

Accepted Answer

Since both have same cross section area , hence stress will be same .For P,  \Delta L=L\alpha_1 T-\dfrac{\sigma L}{Y_1} For Q,  \Delta L=L\alpha_2 T-\dfrac{\sigma L}{Y_2} Let the force be F so  \sigma =F/A For Q,  \Delta L=L\alpha_2 T-\dfrac{F L}{AY_2} Final length of Q = L+\Delta L  = L[1+\alpha_2T-\dfrac{F}{AY_2}]

Question 9

A glass bulb of volume 250cc is filled with mercury at 20 ^{0} C and the temperature is raised to  100^{0}C . If the coefficient of linear expansion of glass is  9	imes 10^{-6/0}C  Coefficient of absolute expansion of mercury is  18	imes 10^{-4/0}C . The volume of mercury that overflows is

Accepted Answer

\gamma_{glass} =3\alpha_{glass}=3 	imes 9 	imes 10^{-6}=27 	imes 10^{-6} \gamma_{mercury} = 18 	imes 10^{-4} Amount of mercury to overflow from the glass  V_{overflow} = V_0 	imes( \gamma_{mercury} -\gamma_{glass} ) 	imes \Delta T= 250 	imes (18 	imes 10^{-4} - 27 	imes 10^{-6}) 	imes (100-20)= 250 	imes 1773 	imes 10^{-6} 	imes  80 \implies V_{overflow}=35.46cc

Question 10

A vessel is half filled with a liquid at 0 ^{0} C. When then vessel is heated to 100 ^{0} C the liquid occupies 3/4 volume of the vessel. Coefficient of apparent expansion of the liquid is :

Accepted Answer

{\gamma}_a=\dfrac{V_2-V_1}{V_1 	imes (t_2-t_1)}  ....(1)Given, V_1=\dfrac{1}{2} V_2=\dfrac{3}{4} t_2-t_1=100-0=100 Putting the values in (1) {\gamma}_a= \dfrac{(\dfrac{3}{4}-\dfrac{1}{2})}{\dfrac {1}{2} 	imes 100}=0.005

Question 11

A vessel contains a liquid filled upto  \dfrac{1}{10} th of its volume. Another vessel contains same liquid upto  \dfrac{1}{8} th of its volume. In both cases volume of empty space remains constant at all temperatures. Then the ratio of the coefficients of the cubical expansion of the two vessels is

Accepted Answer

Increase in volume of vessel = Increase in volume ofliquid {\gamma}_1 	imes V_1 = {\gamma}_2 	imes V_2  	herefore  \dfrac{{\gamma}_1}{{\gamma}_2}= \dfrac{V_2}{V_1}=\dfrac{\frac{1}{10}}{\dfrac{1}{8}}=\dfrac {8}{10}=\dfrac {4}{5}

Question 12

Two metallic spheres  S_{1}  and  S_{2}  are made of the same material and have got identical surface finish. The mass of  S_{1}  is thrice that of  S_{2} . Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. The ratio of the initial rate of cooling of  S_{1}  to that of  S_{2}  is

Accepted Answer

We know that  H=\dfrac{dQ}{dt}=ms\dfrac{d	heta}{dt}  --- (1) H=(4\pi {r}^{2})\sigma {T}^{4}  --- (2)From (1) and (2), we get,  ms\dfrac{d	heta}{dt}=(4\pi {r}^{2})\sigma {T}^{4} Thus,  \dfrac{d	heta}{dt}\propto \dfrac{{r}^{2}}{m} Now,  m=ho 	imes \dfrac{4}{3} \pi {r}^{3}       or,               r \propto {m}^{\dfrac{1}{3}} Thus,              \dfrac{d	heta}{dt}\propto \dfrac{1}{{m}^{\dfrac{1}{3}}} Since mass of  {S}_{1}  is 3 times  {S}_{2} , thus  \dfrac{d	heta}{dt}  for  {S}_{1}  will be  \dfrac{1}{{3}^{\frac{1}{3}}}  times that for  {S}_{2}

Question 13

A hot body is placed in cold surroundings. Its rate of cooling is  3^{o} C per minute when its temperature is  70^{o} C and  1.5^{o} C per minute when its temperature is  50^{o} C. Its rate of cooling when temperature is  40^{o} C is

Accepted Answer

Newton's cooling law:  \dfrac{d	heta}{dt}=-k(	heta-{	heta}_{0}) putting the given values of the two situations in the formula we get:- 3 = -k(70 - {	heta}_{0})  - eq.1 1.5 = -k(50 - {	heta}_{0})  - eq.2Now, solving eq.1 and eq.2 we have  {	heta}_{0} = 30{}^{0}C  and  -k = \dfrac{3}{40} Now solving using the above values for  {40}^{0}C , we have  \dfrac{d	heta}{dt} = \dfrac{3}{40}(40 - 30)  \dfrac{d	heta}{dt}= \dfrac{3}{40}(10) \dfrac{d	heta}{dt}=\dfrac{3}{4} = {0.75}^{0}C/min

Question 14

A body takes  4  minutes to cool from  100^{o} C to  70^{o} C. The time taken by it to cool from  70^{o} C to  40^{o} C in a room with temperature  15^{o} C is nearly_____

Accepted Answer

The Newtons law of cooling formula is given by: T(t) = T_s+(T_o-T_s)e^{-kt} \dfrac{T(t)-T_s}{T_o-T_s}=e^{-kt} -kt=ln\dfrac{T(t)-T_s}{T_o-T_s} -kt=ln\dfrac{70-15}{100-15} Thus we get  k=0.10883 Now for the second interval we get: -kt=ln\dfrac{40-15}{70-15} Thus,  t=7 \ mins

Question 15

A body cools from  100     to  70     in  8   s . If the room temperature is  15     and assuming Newton's law of cooling holds good, then time required for the body to cool from  70     to  40     is:

Accepted Answer

Rate of loss of heat of a body is directly proportional to the difference in temperature of the body and the surroundings.Newton's law of cooling states that the rate of fall of temperature of a body is directly proportional to the difference of temperature between body and surroundings, provided the temperature difference is not more than  30 .According to this law, \dfrac { { 	heta  }_{ 1 }{ -	heta  }_{ 2 } }{ t } = K\left( \dfrac { { 	heta  }_{ 1 }+{ 	heta  }_{ 2 } }{ 2 } -{ 	heta  }_{ 0 } ight)  where  { 	heta  }_{ 0 }=  temperature of surrounding.In first case, \dfrac { 100-70 }{ 8 } =K\left( \dfrac { 100+70 }{ 2 } -15 ight)  or  \dfrac { 15 }{ 4 } =K\left( 70 ight)           .....(i)In second case, \dfrac { 70-40 }{ t } =K\left( \dfrac { 70+40 }{ 2 } -15 ight)  or  \dfrac { 30 }{ t } =K\left( 40 ight)            ......(ii)Dividing equation (i) by equation (ii), we have \dfrac { 15 }{ 4 } 	imes \dfrac { t }{ 30 } =\dfrac { 70 }{ 40 }  t=\dfrac { 7 }{ 4 } 	imes \dfrac { 4	imes 30 }{ 15 }  	herefore t=14s

Thermal Properties Of Matter