Ncert Solution CBSE Class 11 Chapter 10 THERMAL PROPERTIES OF MATTER - Param Himalaya
10.1 The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Solution :
On Celsius scale
t(°C) = T(K) - 273.15
Triple point of neon = 24.57 - 273.15 = - 248.58°C
Triple point of carbon dioxide = 216.55 - 273.15 = -56.6°C
On Fahrenheit scale
$t(^{o}F) = \frac{9} {5} (t^{o}C)+32$
Triple point of neon = $\frac{9} {5} (-248.58)+32$
Triple point of neon= -447.444+32=-415.44°F
Triple point of carbon dioxide = $\frac{9} {5} (-56.6)+32$
Triple point of carbon dioxide = - 101.88 +32 = -69.88°F
10.2 Two absolute scales A and B have triple points of water defined to be 200 A and 350B. What is the relation between $T_{A}$ and $T_{B}$?
Solution :
Triple point of water in
absolute scale A = 200A
absolute scale B = 350 B
Temperature on Kelvin scale = 273.16K
200A = 273.16K
$1A = \frac{273.16}{200}K$
Value of 1 degree on absolute scale A = $\frac{273.16}{200}K$
Value of $T_{A}$ degree on absolute scale A = $\frac{273.16}{200} \times T_{A}$
350B= 273.16K
$1B = \frac{273.16}{350}K$
Value of 1 degree on absolute scale B = $\frac{273.16}{350}K$
Value of $T_{B}$ degree on absolute scale B = $\frac{273.16}{350} \times T_{B}$
As $T_{A}$ and $T_{B}$ represent the same temperature
$\frac{273.16}{200} \times T_{A} =\frac{273.16}{350} \times T_{B}$
$T_{A} = \frac{200}{350}T_{B}$
$\frac{T_{A}}{T_{B}} = \frac{4}{7}$
10.3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law :
R = Ro[1 + α (T – To )]
The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K).What is the temperature when the resistance is 123.4 Ω ?
Solution :
Here $R_{0} = 101.6 \Omega$ , $T_{0}= 273.16 K$
For case (i)
$R_{1} = 165.5 \Omega$
$T_{1} = 600.5 K$
$R_{1} = R_{0}[1+ \alpha(T_{1}-T_{0}]$
$\alpha = \frac{R_{1}-R_{0}}{R_{0}(T_{1}-T_{0})}$
$\alpha = \frac{165.5-101.6}{101.6(600.5-273.16)}$
$\alpha = \frac{63.9}{101.6 \times 327.34}$
$\alpha = \frac{63.9}{33257.744}$
$\alpha = 0.001921 $
$\alpha = 1.92 \times 10^{-3}/K$
Case (ii)
$R_{2} = 123.4 \Omega$ , $T_{2}$= ?
$R_{2} = R_{0}[1+ \alpha(T_{2}-T_{0}]$
$T_{2} = T_{0} + \frac{R_{2}-R_{0}}{\alpha \times R_{o}}$
$T_{2} = 273.16 + \frac{123.4-101.6}{1.921 \times 10^{-3}\times 101.6}$
$T_{2}= 273.16 + \frac{21.6}{195.072}\times 10^{3}$
$T_{2} = 273.16 + 0.11072 \times 10^{3}$
$T_{2} = 273.16 + 110.7$
$T_{2} = 383.8K$
10.4 Answer the following :
(a) The triple-point of water is a standard fixed point in modern thermometry.
Why ? What is wrong in taking the melting point of ice and the boiling point
of water as standard fixed points (as was originally done in the Celsius scale) ?
(b) There were two fixed points in the original Celsius scale as mentioned above
which were assigned the number 0 °C and 100 °C respectively. On the absolute
scale, one of the fixed points is the triple-point of water, which on the Kelvin
absolute scale is assigned the number 273.16 K. What is the other fixed point
on this (Kelvin) scale ?
(c) The absolute temperature (Kelvin scale) T is related to the temperature tc
on
the Celsius scale by
tc
= T – 273.15
Why do we have 273.15 in this relation, and not 273.16 ?
(d) What is the temperature of the triple-point of water on an absolute scale
whose unit interval size is equal to that of the Fahrenheit scale ?
10.5 Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The
following observations are made :
Temperature Pressure Pressure
thermometer A thermometer B
Triple-point of water 1.250 × 105
Pa 0.200 × 105
Pa
Normal melting point 1.797 × 105
Pa 0.287 × 105
Pa
of sulphur
(a) What is the absolute temperature of normal melting point of sulphur as read
by thermometers A and B ?
(b) What do you think is the reason behind the slight difference in answers of
thermometers A and B ? (The thermometers are not faulty). What further
procedure is needed in the experiment to reduce the discrepancy between the
two readings ?
10.6 A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day ?What is the length of the same steel rod on a day when the temperature is 27.0 °C? Coefficient of linear expansion of steel =$1.20 \times10^{–5}C^{-1}$
Solution:
Given $L_{o} = 63 cm$ , $L_{1} = ?$ , $T_{o} = 27^{o}$ , $T_{1} = 45^{o}$
So , $L_{1} = L_{o}[1+ \alpha ( T_{1}-T_{o})]$
$L_{1} = 63[1+ 1.20 \times 10^{-5}(45-27)]$
$L_{1} = 63[1+ 1.20 \times 10^{-5}(18)]$
$L_{1} = 63[1+ 21.6 \times 10^{-5}]$
$L_{1} = 63[1+ 0.000216]$
$L_{1} = 63[1.000216]$
$L_{1} = 63.0136$
length of the same steel rod on a day when the temperature is 27.0 °C will be 63 cm
10.7 A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range :$\alpha_{steel} = 1.20 \times 10^{–5} C^{–1}$.
Solution:
The diameter of the shaft is decreased from 8.70 cm to 8.69 cm using dry ice.
$D_{0} = 8.70 cm$ , $T_{0} = 27^{0}C$
$D_{1} = 8.69 cm$ , $T_{1} = ?$
$D_{1} = D_{0} [1+ \alpha(T_{1}-T_{0})]$
$T_{1}=T_{0} + \frac{D_{1}-D_{0}}{\alpha \times D_{0}}$
$T_{1} = 27 + \frac{8.69-8.70}{ 1.2 \times 10^{-5} \times 8.70}$
$T_{1} = 27 + \frac{-0.01}{10.44 \times 10^{-5}}$
$T_{1} = 27- 0.0009578 \times 10^{5}$
$T_{1} = 27-95.8 = -68.8^{o}C$
10.8 A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C.What is the change in the diameter of the hole when the sheet is heated to 227 °C?Coefficient of linear expansion of copper = 1.70 × 10–5 K–1.
Solution:
When the copper sheet is heated , the diameter of its hole increase in the same manner as the length of a rod.
$\therefore$ Increase in the diameter of the hole is
$\triangle L = L \alpha\triangle T$
Here $L = 4.24 cm , \alpha = 1.70 \times 10^{-5} .^{o}C$
$\triangle T = 227-27 = 200^{o}C$
$\triangle L = 4.24 \times 1.70 \times 10^{-5} \times 200$
$\triangle L = 1.44 \times 10^{-2}cm$
$\triangle L = 0.0144cm$
10.9 A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = $2.0 \times 10^{-5} \ {^\circ}C^{-1}$ ; Young’s modulus of brass = $0.91 \times 10^{11}Pa$.
Solution :
Here $\alpha = 2.0 \times 10^{-5} \ {^\circ}C^{-1}$
$Y =0.91 \times 10^{11}Pa$
Let F be the thermal tension developed in the wire
$Y = \frac{stress}{strain}$
$Y = \frac{\frac{F}{A}}{\frac{\triangle L}{L}}$
$\triangle L = \frac{FL}{AY}$
$F = \frac{AY \triangle L}{L}$
Also $\triangle L = \alpha L \triangle T$
So , $F = \frac{\alpha L \triangle T YA}{L}$
$F = \alpha YA \triangle T$
$A = \pi \times (1.0 \times 10^{-3})^{2}m^{2}$
$A = 3.14 \times 1.0 \times 10^{-6}m^{2}$
$A = 3.14 \times 10^{-6}m^{2}$
$\triangle T = -39-27 = -66 \ ^\circ C$
$F = 2 \times 10^{-5} \times 0.91 \times 10^{11} \times 3.14 \times 10^{-6} \times 66$
$F= 3.77 \times 10^{2}N$
10.10 A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0°C? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = $2.0 \times 10^{-5}{^{\circ}C}^{-1}$, steel = $1.2 \times 10^{-5}{^{\circ}C}^{-1}$.
Solution :
Change in length of the brass rod is
$\triangle L_{b} = \alpha_{b}L \triangle T$
Here $\alpha_{b} = 2.0 \times 10^{-5}/^{\circ}C $
L = 50 cm = 0.5 m
$\triangle T = 250 -40 = 210 ^\circ C$
$\triangle L_{b} = (2.0 \times 10^{-5}) \times (0.5) \times (210)$
$\triangle L_{b}= 2.10 \times 10^{-3} m= 0.21 m$
Change in length of the steel rod is
$\triangle L_{s} = \alpha_{s} L \triangle T$
$\triangle L_{s} = (1.2 \times 10^{-5}) \times (0.5) \times (210)$
$\triangle L_{s} = 1.26 \times 10^{-3}m$
$\triangle L_{s} = 0.13 cm$
Change in length of the combined rod = $\triangle L_{b} + \triangle L_{s} = 0.21 + 0.13 = 0.34 cm$
10.11 The coefficient of volume expansion of glycerine is $49 \times 10^{-5} \ {^{\circ}}C^{-1}$. What is the fractional change in its density for a 30°C rise in temperature ?
Solution :
Given : $\gamma = 49 \times 10^{-5} \ {^{\circ}}C^{-1}$ , $\triangle T$ = 30°C
Let M be the mass of the glycerine and V be it's initial volume . If $V^{'}$ is the volume of glycerin after a rise of temperature of 30°C, then
$V^{'} = V ( 1 + \gamma \triangle T)$
$V^{'} = V ( 1+ 49 \times 10^{-5} \times 30)$
$V{'} = V (1+ \frac{1470}{10^{5}})$
$V{'} = V (1+ \frac{1470}{100000})$
$V{'} = V (1+ 0.01470)$
$V^{'} = V (1.0470)$
$V^{'} = 1.0470 V$
Initial density , $\rho = \frac{M}{V}$
Final density , $\rho{'} = \frac{M}{V^{'}}$
$\rho{'} = \frac{M}{1.0147 V}$
$\rho{'} = \frac{1}{1.0147} .\frac{M}{V}$
$\rho{'} = \frac{1}{1.0147} \rho$
$\rho{'} = 0.9855 \rho$
Fractional change in density
$\frac{\triangle \rho}{\rho} = \frac{\rho - \rho^{'}}{\rho}$
$\frac{\triangle \rho}{\rho} = \frac{\rho - 0.9855 \rho}{\rho}$
$\frac{\triangle \rho}{\rho} =1 - 0.9855$
$\frac{\triangle \rho}{\rho} = 0.0145$
10.12 A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium =$0.91 J g^{-1} \ {^{\circ}}C^{-1}$.
Solution :
Power of the machine ,$P = 10 kW
$P= 10 \times 10^{3}W$
$P= 10^{4}W$
Energy obtained from machine in 2.5 min ,
$E = P.t$
$E = 10^{4} \times 2.5 \times 60 J$
$E = 15 \times 10^{5} J$
Energy transferred to the aluminium block is
Q = 50 % of E
$Q = \frac{50}{100} \times 15 \times 10^{5}$
$Q = 7.5 \times 10^{5}J$
If m is the mass of aluminium block and $\triangle T$ is the rise of its temperature, then
$Q = mc \triangle T$
$\triangle T = \frac{Q}{mc}$
$\triangle T = \frac{7.5 \times 10^{5}}{8.0 \times 10^{3} \times 0.91}$
$\triangle T = 103.02 {^{\circ}}C$
10.13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g–1 K–1; heat of fusion of water= 335 J g–1 ).
Solution :
When copper block remains on ice , it's temperature will decrease from 500°C to 0°C
Heat lost by copper block = mass × specific heat × fall in temperature
$Q_{c}=m_{c}s \triangle T$
$Q_{c}= (2.5 \times 10^{3}) \times 0.39 \times (500-0)$
$Q_{c}= 487500 J$
Let m grams be the mass of ice melted.
Heat gained by ice (latent heat )= mass melted × heat of fusion of water (L)
$Q_{i} = mL$
Heat gained by ice = m × 335 J
But Heat lost = Heat gained
$487500 = m \times 335$
$m = \frac{487500}{335}$
$m = 1455.2 g$
$m = 1.455 kg$
10.14 In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing $150 cm^{3}$ of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ?
Solution :
Mass of metal block ,
$m_{b} = 0.20 kg = 0.20 \times 10^{3}g$
Initial temperature of metal block
$T_{1} = 150 {^{\circ}}C$
Water equivalent of calorimeter ,
$w = 0.025 kg = 0.025 \times 10^{3}g$
Initial temperature of water and calorimeter ,
$T_{2} = 27 {^{\circ}}C$
Final temperature of mixture ,
$T = 40 {^{\circ}}C$
Mass of water = volume × density
$M_{w} = 150 \times 1 = 150 g$
Heat lost by metal block ,
$Q_{b} = m_{b}c (T_{1}-T)$
$Q_{b} = 0.20 \times 10^{3} \times c \times (150-40)$
$Q_{b} = 200 \times c \times 110$
$Q_{b} = 22000c$
Heat gained by Water and Calorimeter ,
$Q_{c} + Q_{w} =(m_{c} + w) \times 4.2 \times (T-T_{2})$
$Q_{c} + Q_{w} = (150+0.025 \times 10^{3}) \times 4.2 \times (40-27)$
$Q_{c} + Q_{w} = (150+25) \times 4.2 \times 13$
$Q_{c} + Q_{w} = 175 \times 4.2 \times 13$
$Q_{c} + Q_{w} = 9555$
So , Heat lost by block = Heat gained by water and calorimeter
$Q_{b} = Q_{c} + Q_{w}$
$22000c = 9555$
$c = \frac{9555}{22000}$
$c= 0.434 J g^{-1}{^{\circ}}C$
10.15 Given below are observations on molar specific heats at room temperature of some common gases.
Gas | Molar specific heat $(C_{v}) (cal mol^{-1}K^{-1})$ |
Hydrogen | 4.87 |
Oxygen | 4.97 |
Nitric oxides | 4.99 |
Carbon monoxide | 5.01 |
Chlorine | 6.17 |
The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ?
10.16 A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 minutes, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1.
Solution :
Given:
The initial temperature of the child, $T_{i}=101{^{\circ}}F$
The final temperature of the child, $T_{f}=98{^{\circ}}F$
Decrease in the temperature,
$\triangle T =(101−98)= 3{^{\circ}}F$
$\triangle T = 3 \times \frac{5}{9}= 1.670{^{\circ}}C$
Mass of the child,
$m=30 kg=30 \times 10^{3}g$
Time is taken to reduce the temperature, t = 20 min
Specific heat of the human body = Specific heat of water
$c =1 cal g^{-1}{^{\circ}}C^{-1}$
Latent heat of evaporation of water,
$L=580 cal g^{−1}$
Calculation:
The following is the amount of heat lost by the child:
Q=mc△T
Q=30000×1 × 1.67
Q=50100 cal
Let m’ be the amount of water evaporated from the child’s body .
Q= m'L
$m^{'} =\frac{Q}{L}$
$m^{'} = \frac{50100}{580}$
$m^{'} = 86.37g$
Therefore, the average rate of evaporation=86.2/20 =4.3 g/min.
10.17 A ‘thermacole’ icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is $0.01 J s^{–1}m^{–1}K^{–1}$. [Heat of fusion of water = $335 \times 10^{3}J kg^{–1}$]
Solution :
Here
$k = 0.01 Js^{-1}m^{-1} {^{\circ}}C$
$A = 30 \times 30 = 0.3 \times 0.3 m^{2}$
$A = 0.09 m^{2}$
$\triangle T = 45{^{\circ}} - 0{^{\circ}} = 45 {^{\circ}}$
$t = 6 \times 60 \times 60 = 21.6 \times 10^{3}sec$
$x = 5.0 cm = 5.0 \times 10^{-2}m$
The heat entering the box through all its six faces in 6 h is
$Q = 6 \times \frac{kA \triangle Tt}{x}$
$Q = 6 \times \frac{0.01 \times 0.09 \times 45 \times 21.6 \times 10^{3}}{5 \times 10^{-3}}$
$Q=104976J$
Suppose m kg of ice melts in 6 hours.
Now , Q = mL
$m = \frac{Q}{L}$
$m = \frac{104976}{3.35 \times 10^{3}}$
$m = 0.3 kg$
Amount of ice remaining = 4-0.3 = 3.7 kg
10.18 A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = $10^{9}J s^{–1}m^{–1}K^{–1}$,Heat of vaporisation of water = $2256 \times 10^{3}J kg^{–1}$
Solution:
Base area of the boiler , $A = 0.15 m^{2}$
Thickness of the boiler , l = 1.0 cm = 0.01 m
Boiling rate of water , $R = 6.0 \frac{kg}{min}$
Mass , m = 6kg
Time , t = 1 min = 60 s
Thermal conductivity of brass
$K= 10^{9}J s^{–1}m^{–1}K^{–1}$
Heat of vaporisation , $L = 2256 \times 10^{3}J kg^{–1}$
The amount of heat flowing into water through the brass base of the boiler is given by
$Q = \frac{kA \triangle T.t}{l}$
where $T_{1}$= Temperature of the flame in contact with the boiler
$T_{2}$ = Boiling point of water = 100°C
Heat required for boiling the water
$Q = mL$
So
$mL = \frac{kA \triangle T.t}{l}$
$\triangle T = \frac{mLl}{KAt}$
$\triangle T = \frac{6 \times 2256 \times 10^{3} \times 0.01}{10^{9} \times 0.15 \times 60}$
$\triangle T = 137.98{^{\circ}}C$
$T_{1} - T_{2} = 137.98{^{\circ}}C$
$T_{1} = T_{2} + 137.98{^{\circ}}C$
$T_{1} = 100{^{\circ}}C + 137.98{^{\circ}}C$
$T_{1} = 237.98{^{\circ}}C$
10.19 Explain why :
(a) a body with large reflectivity is a poor emitter
(b) a brass tumbler feels much colder than a wooden tray on a chilly day
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace
(d) the earth without its atmosphere would be inhospitably cold.
(e) heating systems based on circulation of Steam are more efficient in warming a building than those based on circulation of hot water.
10.20 A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.
Solution:
Here , Initial temperature $T_{i} = 80 {^\circ}C$
Final temperature $T_{f} = 50 {^\circ}C$
Temperature of the surrounding ($T_{0})= 20 {^\circ}C$
t = 5 min
According to Newton's law of cooling
Rate of cooling $\frac{dT}{dt} = K [\frac{(T_{i}+T_{f})}{2} - T_{o}]$
$\frac{(T_{f}-T_{i})}{t} = K[\frac{80+50}{2}-20]$
$[\frac{80-50}{5}] = K[65-20]$
$6 = K \times 45$
$K = \frac{6}{45}$
$K = \frac{2}{15}$
In second condition
Initial temperature $T_{i} = 60 {^\circ}C$
Final temperature $T_{f} = 30 {^\circ}C$
Time taken for cooling is t
According to Newton's law of cooling
$\frac{(60-30)}{t} = \frac{2}{15}[\frac{60+30}{2}-20]$
$\frac{30}{t} = \frac{2}{15} \times 25$
$\frac{30}{t} = \frac{50}{15}$
$\frac{30}{t} = \frac{10}{3}$
$t = 9 min$