Thermal Properties of Matter

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Decrease in temperature ? t = 57-37= 200C

Coefficient of linear expansion α = 1.7 * 10 - 5 / oC

Bulk modulus for copper B = 140 * 10 9 N / m 2

Coefficient of cubical expansion γ = 3 α = 5.1 * 10 - 5 / ° C

Let initial volume of the cavity be V and its volume increases by ? V  due to increase in temperature.

? V = γ V ? t

? V V = γ ? t

Thermal stress produced = B * v o l u m e t r i c s t r a i n

= B * ? V V = B γ ? t

= 140 * 10 9 * 5.1 * 10 - 5 * 20

= 1428 * 10 8 N / m 2

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4 months ago

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

As difference in volume is constant

By considering the diagram

Let Vio , Vbo be the volume of iron and brass vessel at 00C

Vi,Vb be the volume of iron and brass vessel at ? θ 0C

γ i , γ b be the coefficient of volume expansion of iron and brass.

Vio -Vbo= 100cc= Vi-Vb

Vi =Vio(1+ γ i ? θ )

Vb =Vbo(1+ γ b ? θ )

Vi-Vb = (Vio -Vbo)+ ? θ V i o γ i - V b o γ b

Since Vi-Vb= constant

Vio γ i= Vbo γ b

V i o V b 0 = γ b γ i = 3 2 β b 3 2 β i = β b β i = 6 * 10 - 5 3.55 * 10 - 5 = 6 3.55

Using above equations

V i o = 244.9 c c

Vbo = 144.9cc

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4 months ago

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

As liron-lbrass =10cm= constant at all temperature

Let lo be the length of temperature at 00C  and l be the length after change in temperature

liron-lbrass =10cm

liron (1+ α i r o n ? t )-lbrass (1+ α b r a s s ? t )=10cm

Iiron α iron= Ibrass α brass

l i r o n l b r a s s =1.8/1.2=3/2

1 2 l b r a s s = 10 c m

Lbrass=20cm and liron=30cm

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Payal Gupta

Contributor-Level 10

11.22

The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporization curves. It departs from ideal gas behavior as pressure increases.

(a) At 1 atm pressure and at – 60 °C, it lies left of -56.6 ? (triple point O). Hence, it lies in the region of vapour and solid phases. Thus,  CO2  condenses into solid state directly, without going through liquid phases.

(b) At 4 atm pressure,  CO2 lies below 5.11 atm (triple point O). Hence it lies in the region of vaporous and solid phases. Thus, CO2 condenses into solid state directly, without going through liquid state.

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Payal Gupta

Contributor-Level 10

11.21 

(a) The P-T phase diagram for CO2 is shown here. O is the triple point of the CO2 phase diagram. This means that at the temperature and pressure corresponding to this point, the solid, liquid and vaporous phases of CO2 exists in equilibrium.

(b) The fusion and the boiling points of CO2 decreases with a decrease in pressure.

(c) The critical temperature and critical pressure of CO2 are 31 ? 72.9 atm respectively. Even if it is compressed to a pressure greater than 72.9 atm, CO2 will not liquefy above the critical temperature.

(d) It can be concluded from the P-T phase diagram of CO2 that:

CO2 is gaseous at -70 ? , under 1 atm pr

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Payal Gupta

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11.20 According to Newton’s law of cooling, we have -dTdt
 = K(T - To ) or dTK(T-To) = -Kdt……(i)Where, temperature of the body = T

Temperature of the surroundings = To = 20 °C

K is a constant

The temperature of the body falls from 80? to 50?intimet=5min?=300s

Integrating equation (i), we get

5080dTK(T-To) = - 0300Kdt∫

 ?loge?(T-To)3080= -K t0300

 ?2.3026Klog10?80-2050-20= -300

?2.3026300log10?2 = K ….(ii)

The temperature of the body falls from 60 ? to 30 ?intime=t'

Hence, we get:

2.3026Klog10?60-2030-20 = -t'


?-2.3026t'log10?4 = K.(iii)

Equating equations (ii) and (iii), we get,


2.3026300log10?2= -2.3026t'log10?4

?t'=300*2=600s = 10 min

Hence, the time taken to cool the body from 60 ? to 30 ? is 10 minutes.

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4 months ago

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Payal Gupta

Contributor-Level 10

11.19 (a) A body with a large reflectivity is a bad absorber. A bad absorber will in turn be a poor emitter of radiations.

(b) Brass is a good conductor of heat, when one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a colder value and one feels cold.

On the other hand, wood is a poor conductor of heat. Very little heat is conducted from the body to the wooden tray. Resulting in negligible drop in body temperature.

Thus a brass tumbler feels colder than a wooden tray on a chilly day.

(c) Black body radiation equation is given by:

E = σ ( T4&nbs

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4 months ago

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Payal Gupta

Contributor-Level 10

11.18 Base area of the boiler, A = 0.15 m2

Thickness of the boiler, l = 1.0 cm = 0.01 m

Boiling rate of water, R = 6.0 kg/min

Let us assume the mass of the boiling water, m = 6 kg and the time to boil, t = 1 min = 60 s

Thermal conductivity of brass, K = 109 J s–1 m–1 K–1

The amount of heat flowing into water through the brass base of the boiler is given by:

θ = KAT1-T2tl , where

T1 = Flame temperature in contact with the boiler

T2 = Boiling point of water = 100 ?

Heat required for boiling water θ = mL, where L = heat of vaporization of water = 2256 * 103 J kg–1

By equating for θ we get

*2256*103 = 109&n

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4 months ago

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Payal Gupta

Contributor-Level 10

11.17 Size of the sides of cubical ice box, s = 30 cm =0.3 m

Thickness of the icebox, l = 5 cm = 0.05 m

Mass of ice kept in the box, m = 4 kg

Time gap, t = 6 h = 6 *60*60 s

Outside temperature, T = 45 °C

Coefficient of thermal conductivity of thermocole, K = 0.01 J s-1m-1K-1

Heat of fusion of water, L = 335 *103 J kg-1

Let m' be the mass of the ice melts in 6 h

The amount of heat lost by the food: θ = KAT-0tl , where

A = Surface area of the box = 6 s2 = 6 *0.32m2 = 0.54 m2

θ = 0.01*0.5445-0*6*60*600.05 = 104976 J

We also know θ=m'L so m' = 104976/ (335 *103) = 0.313 kg

Hence the amount of ice remains after 6 h = 4

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4 months ago

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Payal Gupta

Contributor-Level 10

11.16 Initial body temp of the child,  T1 = 101°F

Final body temp of the child,  T2 = 98°F

Change in temperature,  ?  T = T1-T2= (101°F- 98°F) = 3 °F = 59  (3-32) ° C = 1.666 ?

Time taken t achieve this temperature, t = 20 min

Specific heat of human body = Specific heat of water, c = 1000 cal/kg/ ?

Latent heat of evaporation of water, L = 580 cal/g

Mass of the child, m = 30 kg

The heat lost by the child is given as ? θ=mc? T = 30 *1000*1.666 = 49980 cal

Let m1 be the mass of water evaporated from the child's body in 20 mins.

Loss of heat ? θ = m1*L = 580 m1 = 49980cal

m1=&nb

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