Q.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.
Ans:
Given: Kelvin and Celsius scales are related as
TC = TK – 273.15 . . . . . . . . . . . . . . (1)
We know :
TF =
Q.2: Two absolute scales, A and B, have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB?
Ans:
Given:
The triple point of water on absolute scale B, T2 = 400 B
The triple point of water on absolute scale A, T1 = 200 A
The triple point of water on the Kelvin scale, TK = 273.15 K
273.15 K on the Kelvin scale is equivalent to 200 A on the absolute scale A.
Q.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?
Ans:
Triple point temperature, T0 =273.16 K
Resistance at the triple point, R0=101.6 Ω
The normal melting point of lead, T1= 600.5 K
Resistance at the normal melting point, R1 =165.5 Ω
According to approximate law,
R1=R0[1+α(T1−T0)]
165.5=101.6[1+α(600.5 – 273.16)]
165.5=101.6[1+α(327.34)]
165.5=101.6+α(101.6)(327.34)
165.5=101.6+α(101.6 x 327.34)
α = (165.5 – 101.6) /(101.6 x 327.34)
α = 63.9/(101.6 x 327.34)
Now, when resistance is 123.4Ω, then temperature T2 is
R2=R0[1+α(T2−T0)] 123.4=101.6[1+α(T2 – 273.16)]
123.4= 101.6[1 +(63.9/(101.6 x 327.34)) (T2 – 273.16)]
123.4 = 101.6 + (63.9/327.34)(T2 – 273.16)
123.4 =101.6 + (0.195)(T2) – (0.195) (273.16)
123.4 =101.6 + (0.195)(T2) – 53.32
T2 = (123.4 – 101.6 + 53.32) /0.195 = 75.12/0.195 = 385.23
Q.4: Answer the following.
(a) The triple-point of water is a standard fixed point in a modern thermometer. Why? What is wrong with 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 numbers 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?
Ans:
(i) Melting and boiling points of water aren’t considered the standard fixed points because they vary with change in pressure, the temperature of the triple point of water is unique, and it does not vary with pressure.
(ii) On Kelvin’s scale, there is only a lower fixed point which is 273.16 K, the upper fixed point is not there.
(iii) The relation is such because 273.15 K on Kelvin’s scale corresponds to the melting point of ice while 273.16 K is the triple point of water.
(iv) We know,
Relation between the Fahrenheit scale and the Absolute scale,
i.e.
Therefore, Fractional change in density = 1.47 × 10-2
Q.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 K–1.
Ans:
Power = 10 kW
Mass of the small aluminium block, m = 8 kg = 8 x 103 g
Time = 2.5×60=150 s
Specific heat of aluminium, c= 0.91 J g–1 K–1
Total energy =P×t=104 ×150=15×105 J
As 50% of the energy is used in the heating or lost to the surrounding,
Therefore, thermal energy available, ΔQ= (1/2) x 15×105
= 7.5 x 105
As ΔQ=mcΔT
ΔT= ΔQ/mc
= 7.5 x 105/ (8 x 103 ×0.91)
= 7.5 x 105/ 7.28 x 103
= 1.03 x 102
Rise in the temperature of the block, ΔT = 1030 C
Q.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; the heat of fusion of water = 335 J g–1 )
Ans:
The mass of the copper block, m = 2.5 kg
The temperature of the block, ΔT= 5000 C
The pecific heat of copper, c= 0.39 J g–1 K–1
The latent heat of fusion of water, L = 335 J g–1
Let m’ be the mass of the ice melted.
Therefore, Heat gained by ice = Heat lost by copper
m’L = mcΔT
m’ = mc ΔT/L
= (2500 x 0.39 x 500)/335 = 1500 g = 1.5 kg
Q.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 cm3 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 the specific heat of the metal?
Ans:
The mass of the metal block, m = 0.20 kg = 200 g
The initial temperature of the metal block, = 1500C
The final temperature of the metal block, = 400C
The copper calorimeter has water equivalent of mass, m1= 0.025 kg = 25 g
The volume of water, V = 150 cm3
The mass (M) of water at temperature T = 270C is
Specific heat of water, Cw
Specific heat of the metal = c
Decrease in the temperature of the metal block
–
Increase in the temperature of the water and calorimeter system,
Heat lost by the metal = Heat gained by the water + Heat gained by the calorimeter
mCΔT1 = (M+m1)Cw
C = [(M+m1)Cw]/mΔT1
= [(150 +25) x 4.186 x 13]/(200 x 100)
=(175 x 4.186 x 13)/(200 x 100)
= 9523.15/22000
= 0.43
If some heat is lost to the surroundings, the specific heat of metal will be lesser than the actual value.
Q.15: Given below are observations on molar specific heats at room temperature of some common gases.
| Gas | Molar Specific Heat CV ( cal mol-1 K-1) |
| Chlorine | 6.17 |
| Oxygen | 5.02 |
| Carbon monoxide | 5.01 |
| Nitric oxide | 4.99 |
| Nitrogen | 4.97 |
| Hydrogen | 4.87 |
Generally, the specific heat of a monoatomic gas is 2.92 cal (mol K )‑1 , which is significantly lower than the specific heat of the above gases. Explain. It can be observed that chlorine has little larger value of specific heat, what could be the reason?
Ans:
The gases in the above list are all diatomic, and a diatomic molecule has translational, vibrational and rotational motion. Whereas a monoatomic gas only has translational motion. So, to increase the temperature of one mole of a diatomic gas by 1°C, heat needs to be supplied to increase translational, rotational and vibrational energy. Thus, the above gases have significantly higher specific heat than monoatomic gases Chlorine has a little larger specific heat as compared to the others in the list because it possesses vibrational motion as well, while the rest only have rotational and translational motions.
Q.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 the human body is approximately the same as that of water, and the latent heat of evaporation of water at that temperature is about 580 cal g–1.
Ans:
The initial temperature of the child, Ti =101o F
The final temperature of the child, Tf=98oF
The decrease in the temperature, △T=(101−98)= 30 F = 3×(5/9)= 1.670 C
The mass of the child, m=30 kg=30×103 g
The time taken to reduce the temperature, t = 20 min
The specific heat of the human body = Specific heat of water = c =1000 cal kg-1 0C-1
The latent heat of evaporation of water, L=580 cal g−1
The heat lost by the child is given as
△θ=mc△T
=30×1000× 1.67
=50100 cal
Let m’ be the amount of water evaporated from the child’s body in 20 min.
m’ =△θ/L
=(50100/580)=86.37 g
Therefore, the average rate of evaporation=86.2/20
=4.3 g/min.
Q. 17: A ‘thermacole’ icebox is a cheap and 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 the co-efficient of thermal conductivity of thermacole is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 × 103 J kg–1]
Ans:
The side of the cubical icebox, s =30 cm=3 x 10-2 m
The thickness of the icebox, L =5.0 cm=0.05 m
The mass of ice kept in the icebox, m=4 kg
Time, t=6 h=6×60×60 = 21600
Outside temperature, T1= 45o C
The temperature of the icebox = 00 C
The temperature difference = T1 – T2 =45o C – 00 C
The surface area of the icebox = 6 x (0.30)2= 0.54
The coefficient of thermal conductivity of thermacole, K=0.01 Js−1m−1k−1
The heat of fusion of water, L= 335×10 3 Jkg −1
The total heat entering the icebox in 6 hours is
Q = KA(T1−T2)t/L
= (0.01 Js-1m-1C-1 x 0.54 m2 x 450 C x 21600 s)/0.05 m
= 1.05 x 105 J
Let m be the total amount of ice that melts in 6 h.
But Q= mL
Therefore, m = Q/L
=1.05 x 105 /(335×103)=0.313 kg
Amount of ice remaining after 6 h = 4–0.313=3.687 kg
Q.18.A brass boiler has a base area of 0.15 m2 and a thickness of 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 = 109 J s–1 m–1 K–1; Heat of vaporisation of water = 2256 × 103 J kg–1.
Ans:
The base area of the brass boiler, A= 0.15 m2
The thickness of the boiler, d =1.0 cm=0.01 m
The brass boiler boils water at the rate, R=6.0 kg/min
Time, t = 60 sec
Mass = 6 kg
The thermal conductivity of brass, K =109 Js−1m−1k−1
The heat of vaporisation of water, L = 2256×103 Jkg−1
Let T1 be the temperature of the flame in contact with the boiler.
Let T2 be the boiling point of water = 100 0C
Let Q be the amount of heat flowing into the water through the base of the boiler.
Q =KA(T1 −T2)t/d
Q = [109 x 0.15 x (T1 – 100) x 60]/ 0.01 ——-(1)
Heat received by water
Q=mL
Q = 6 x 2256×103 ——-(2)
Equating equations (1) and (2), we get
mL = KA(T1 −T2)t/d
6 x 2256×103 = [109 x 0.15 x (T1 – 100) x 60]/ 0.01
13536 ×103 x 0.01 = 981 (T1 – 100)
T1 – 100 = 13536 ×103 x 0.01/ 981
T1 = 137.9 + 100
= 237. 9 oC
Therefore, the temperature of the part of the flame in contact with the boiler is 237.98 oC.
Q.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 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 the circulation of steam are more efficient in warming a building than those based on the circulation of hot water.
Ans:
(a) A body with a large reflectivity is a poor absorber of heat radiation. A poor absorber will be a poor emitter of radiation. Therefore, a body with a large reflectivity is a poor emitter.
(b) Brass is a good conductor of heat, and wood is a poor conductor of heat. When we touch a brass tumbler, heat is conducted from our hand to the tumbler, and there is a drop in the body temperature; therefore, we feel cold. When a wooden tray is touched on a chilly day, very less heat is conducted from the hand to the wooden tray, and body temperature is not decreased much. So we do not feel cold.
(c) The radiation energy from a red-hot iron piece placed in a furnace is given by the relation
E=σT4
When the iron piece is placed in the open, the radiation energy is given by the relation
E=σ(T4−T04)
Here, E is the energy radiation.
T is the temperature of the optical pyrometer.
σ is a constant.
T0 is the temperature of the open space.
The increase in the temperature of the open space reduces the radiation energy.
(d) If there is no atmosphere, extra heat will not be trapped. All the heat from the sun will be radiated back from the surface of the earth. So without the atmosphere, the earth would be inhospitably cold.
(e) Steam at 100 0C is much hotter than water at 100 0C. This is because steam will have a lot of heat in the form of latent heat (540 cal/g). Therefore, heating systems based on the circulation of steam are more efficient in warming a building than those based on the circulation of hot water.
Q.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.
Ans:
Here,
The initial temperature of the body, T1=80 0C
The final temperature of the body, T2 = =50 0C
Average temperature, (T1 + T2) /2
= (80 + 50)/2 = 65 0C
The temperature of the surrounding, T0 =20 0C
The temperature difference, ΔT= 65 0C – 20 0C = 45 0C
t=5 min.
According to Newton’s law of cooling, the rate of cooling is proportional to the difference in temperature
Change in temperature/ Time = K ΔT
(T2 – T1)/t = KΔT
(80 – 50)/5 = K(45)
6 = K(45)
K = 6/45
K = 2/15
In the second condition,
The initial temperature of the body, T1 =60 0C
The final temperature of the body, T2 =30 0C
Let t be the time taken for cooling.
Average temperature, (T1 + T2)/2 = (60 + 30)/2 = 45 0C
Temperature difference, ΔT = 45 0C – 20 0C = 25 0C
According to Newton’s law of cooling, the rate of cooling is proportional to the difference in temperature
Change in temperature/ Time = K ΔT
(T2 – T1)/t = KΔT
(60 – 30)/t = (2/15) (25)
30/t = 3.33
t = 30/3.33 = 9 min
Q.21. Answer the following questions based on the P-T phase diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium?
(b) What is the effect of the decrease of pressure on the fusion and boiling point of CO2?
(c) What are the critical temperatures and pressure for CO2? What is its significance?
(d) Is CO2 solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm, (c) 15 °C under 56 atm?
Ans:
(a) Liquid and vapour phases of CO2 co-exist at the triple point temperature = – 56.6 °C and pressure = 5.11 atm.
(b) Both the boiling point and freezing point of CO2 decrease if pressure decreases.
(c) The critical temperature and pressure of CO2 are 31.1 °C and 73.0 atm, respectively.
Above this temperature, CO2 will not liquefy, even if compressed to high pressures.
(d) (a) vapour (b) solid (c) liquid
Q.22: A hot ball cools from 90 °C to 10 °C in 5 minutes. If the surrounding temperature is 20°C, what is the time taken to cool from 60 °C to 30 °C?
Ans:
Using Newton’s law of cooling, the cooling rate is directly proportional to the difference in temperature.
Here, average of 90C and 40 °C = 50 °C
Surrounding temperature = 20 °C
Difference = 50 – 20= 30° C
Under the given conditions, the ball cools to 80° C for 5 minutes
Therefore,
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