# thermal properties of matter important problems

Question 1
the triple points of neon and carbon dioxide are 24.57 k and 216.55K respectively. Express these temperatures on Celsius and Fahrenheit scales

Triple point of neon = 24.57 K
and, triple point of CO2 = 216.55K
these two temperatures are given on absolute or Kelvin temperature scale.
On Celsius Scale:-
TC = TK - 273
Where TC is temperature on Celsius scale
TK is temperature on Kelvin scale
Thus,
Triple point of neon = 24.57 - 273.15 = -248.58°C
Triple point of C02 = 216.55 - 273.15 = -56.6°C
On Fahrenheit scale:-
TF = (9TC/ 5 )+ 32
TF is temperature on Fahrenheit scale.
TC is temperature on Celsius scale
Thus,
Triple point of neon = (9/5)(-248.48) + 32
= 415.44 °F
Triple point of CO2 = (9/5)(-56.6) + 32
= - 69.88 ° F

Question 2
Two absolute scales A and B have triple points of water defined as 200A and 350A. What is the relation between TA and TB

Given that on absolute scale
Triple point of water on scale A = 200 A
Triple point of water on scale B = 350 B
Also, triple point of water on Kelvin scale = 273.16 K
Now temperature on scale A and on scale B is equivalent to 273.16 K on absolute temperature scale.
Thus, value of one degree on absolute scale A = (273.16/200) K
Or,
Value of temperature TA on absolute scale A = (273.16XTA)/200
Similarly value of temperature TB on absolute scale B = (273.16XTB)/350
Since TA and TB represent the same temperature
273.16×TA/200 = 273.16×TB/350
Or, TA = 200TB/350 = 4TB/ 7

Question 3
The pressure of the gas in constant volume gas thermometer are 80 cm, 90 cm and 100 cm of mercury at the ice point, the steam point and in a heated wax bath respectively. Find the temperature of the wax bath.

Given that,
Pressure at ice point Pice = 80 cm Hg
Pressure at steam Psteam = 90 cm Hg
Pressure at heated wax bath Pwax = 100 cm Hg
The temperature of wax bath on the scale measured by thermometer is
$T_{wax} =\frac {P_{wax} - P_{ice}}{P_{steam} - P_{ice}} \times 100$ ° C
$=\frac {100cm Hg - 80 cm Hg}{90cm Hg - 80 cm Hg} \times 100$ °C
=200 ° C

Question 4
A resistance thermometer reads R= 20.0Ω, 27.5Ω and 50.0 Ω at the ice point (0°C), the steam point (100°C) and the zinc point 420°C respectively. Assuming that the resistance varies with temperature as Rθ=R0(1+αθ+βθ2), find the values of R0, α and β. Here θ represents temperature on Celsius scale.

Given are the values of resistances,
at ice point 0°C, R0 = 20.0Ω
at steam point 100°C, R100 = 27.5Ω
at zinc point 420°C, Rz = 50.0Ω
Now resistance Rθ varies as
Rθ = R0 (1+αθ+βθ2)
For resistance at ice point
Rice= R0 (1 +α°C +β(0°C)2)
Rice = R0 = 20°C
For resistance at steam point
Rsteam= R0 (1 +α(100°C) +β(100°C2)
27.5α / 20.0α = 1 + 100 (α+ 100β)
or,
α+100β= 3.75 × 10-3 .......(i)
for resistance at zinc point
Rz = R0 [1 +α(420°C) +β(420°C)2]
(50Ω/ 20.0Ω)- 1 = 420 (α+420β)
or α+ 420β= 3.75 × 10-3 .......(ii)
Subtracting (i) and (ii) we find the value of β
α+ 420 β- α-100 β= 3.57 × 10-3-3.75× 10-3
or
320β= - 0.18× 10-3
or,
β = - 5.62× 10-3 °C-1
Put the value of β in eqn (i) for finding value of α
thus α= 3.75× 10-3- 100 (-5.62× 10-7)
or,
α= 3.8× 10-3 °C-1

Question 5
A circular hole of diameter 2.00 cm is made in an aluminium plate at 0°C. What will be the diameter at 100°C. α for aluminium = 2.3 × 10-3 °C-1.

Diameter of circular hole in aluminium plate at 0°C = 2.0 cm
With increase in temperature from 0°C to 100°C diameter of ring increases.
Using,
l = l0(1 +αΔθ)
Where,
l0 = 2.0 cm
α= 2.3 x 10-3 °C-1
and Δθ= (100°C - 0°C) = 100°C
we can find diameter at 100°C
l= 2(1 + 2.3 x 10-5X 100°C)
= 2.0046 cm.

Question 6
The density of water at 0°C is 0.998 gm/cm3 and at 4°C is 1.000 gm/cm3. Calculate the average coefficient of volume expansion of water in temperature range 0 to 4°C.

Density is mass per unit volume given in the question is density of water at 0°C and 4°C
ρ0 = .998 gm/cm3 and ρ4=1.00 gm/cm3 with increase in temperature volume of water changes using,
V = V0 (1 +βθ)
β is the coefficient of volume expansion.
Volume at 4°C equals
V4 = V0(1 +β (4°C - 0°C))
or V4 = V0 (1 + 4β)
If m is the mass of water then density of water at 4°C is
m/V4 =m/V0 [(1 + 4β)]
or ρ4 = ρ0 / (1 +4β)
1.00 =.998/(1+4β)
1 +4β = .998
β = -5 X 10-4 °C-1

Question 7
Refer to the below plot of temperature versus time showing the changes in the state of ice on heating (not to scale)

Which of the following is correct?
(a) The region AB represents ice and water in thermal equilibrium.
(b) At B water starts boiling.
(c) At C all the water gets converted into steam.
(d) C to D represents water and steam in equilibrium at boiling point.

(a) and (d)

Question 8
A glass vessel measures exactly 10cm × 10cm × 10cm at 0°C. It is filled completely with mercury at this temperature. When the temperature rises to 10°C , 1.6 cm3 of mercury overflows. Calculate the Coefficient of volume expansion of mercury. Coefficient of linear expansion of glass = 6.5 × 10-6°C-1.

At temperature θ= 0°C Volume of glass vessel is
V = 1000 cm3
It is then completely filled with mercury.
At temperature θ= 10°C, 1.6 cm3 of mercury overflows
we have to calculate coefficient of volume expansion of mercury γm
Coefficient of volume expansion of glass γg= 3α
γg = 3 × 6.5 × 10-6 °C-1
= 1.95 × 10-5 °C-1
Volume expansion of mercury at 10°C
V=V0 (1+γmΔθ)
and that of glass.
V=V0 (1+γgΔθ)
now volume of mercury overflown is
V-V = V0mg)Δθ
1.6 cm3 = 1000 × 10°C (γm - 1.95 × 10-5)
γm = 1.795 × 10-4
or approximately
γm = 1.8 × 10-4 °C-1.

Question 9
A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at 20°C. Find the longitudinal strain developed in the rod if temperature rises to 50°C. Coefficient of linear expansion of steel is 1.2 × 10-5°C-1

Due to increase in temperature there would be an fractional increase in length of the rod. But since the rod is rigidly clamped. The tension must increase by sufficient amount to produce and equal and opposite change in length. Thus total change in length due to thermal expansion plus longitudinal strain is zero
αΔθ+longitudinal strain = 0
longitudinal = -αΔθ
= - 1.2 × 10-5 × (50°C - 20°C)
= - 3.6 × 10-4

Question 10
To ensure a light fit aluminium reverts used in air-plane construction are made slightly larger than the revert holes and cooled by dry ice (solid CO2) before being driven. If diameter of a hole is 0.2500 in, what should be the diameter of revert at 20°C if is diameter is equal to that of hole when revert is cooled to -78°C the temperature of dry ice? Coefficient of Linear expansion of Aluminium =2.4X10-5°C-1

Given the diameter of hole = 0.2500 inch
at -78°C temperature of dry ice diameter of revert Dr = 0.2500 in
To find diameter of revert at 20°C using,
Dr20 = Dr (1 + αal Δθ)
= 0.2500 (1 + 2.4 × 10-5(20-(-78))
= .2506 inch

Question 11
A steel ring of 3.000 in inside diameter at 20°C is to be heated and slipped over a brass shaft measuring 3.002 inch in diameter at 20°C αbrass = 2 × 10-5 °C-1 αsteel =1.2 × 10-5 °C-1
(a) To what temperature should the ring be heated?
(b) If the ring and shaft together are cooled by some means such as liquid air, at what temperature will the ring just slip off the shaft.

At 20°C
(a) Diameter of brass shaft is 3.002 inch and inner diameter of steel ring is 3.000 inch. Steel ring is to be slipped into the brass shaft and is required to be heated till it fits the shaft. Using equation,
Dθ = D20 (1+ αSΔθ)
3.002 = 3.000(1+1.2 × 10 -5 (θ-20°C))
Calculating for value of θ we finally get
θ = 75.5°C
(b) Now we have to find the temp until which system is to be cooled so that ring slips off the shaft.
If D and D be the diameter of ring and shaft and are equal at temperature θ, then D= D=D
Using,
(i) for ring
D= 3[1+αs(θ- 20)]
(ii) for shaft
D= 3.002[1+αb(θ- 20)]
Subtracting (i) from (ii)
3.002+αb × 3.002(θ- 20) - 3-3αs(θ- 20) = 0
or ,
(θ- 20) =- .002/[(3.002 X 2 X 10-5) - (3 × 1.2 × 10-5)]
θ= - 0.002/(2.404 × 105 + 20)
= - 63.19°C.

Question 12
Specific heat capacity of the a metal A is expressed
as
C=K1T+K2T3
How much heat energy is required to raise the temperature from T0 to 2T0 of unit mass of the metal A
a. (T02/4)(6K1 +15K2T0)
b. (T02/4)(6K1 +15K2T02)
c. (T02/4)(6K1 -15K2T02)
d. None of the above

dQ=CdT
Integrating with limits T0 from 2T0
∫dQ=∫(K1T+K2T3)dT
Q=(T02/4)(6K1 +15K2T02)

Question 13
A aluminium container of mass .65 kg contains .35kg of water at 18° C.A block of iron of mass .2kg at 100° C is put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium ,iron and water are 910 J/Kg-K,470J/kg-K and 4200 J/kg-K respectively.

Mass of aluminium container=.65 Kg
Mass of water=.35 Kg
Mass of iron block=.2 Kg

Temperature of iron block=100° C
Temperature of aluminium water system=18° C

Now heat gained by water
Q1=.35*4200(T-18)
Heat gained by the aluminium
Q2=.65*910(T-18)

Heat lost by iron block
Q3=.2*4700(100-T)
Now
Q1+Q2=Q3

Substituting all values and solving
T=21.58° C

Question 14
An iron piece of mass 200 gm is kept inside a furnace for a considerable long time and then put in calorimeter of water equivalent 20 Gm containing 280 gm of water at 22° C. The mixture attains an equilibrium temperature of 80° C. Find the temperature of the furnace.
Specific heat capacities of iron 470 J/kg-° C.

Mass of iron piece= 200 × 10-3 Kg
mass of water and calorimeter= 300 × 10-3 Kg
Heat gained by calorimeter
Q=msΔθ
s=4186 J/kg-K is the specific heat capacity of water.
thus
Q1=300 × 10-3 Kg × 4186 J/kg-K (80° C - 22° C)
or Q1=1255.8 J/K × 58° C
now, heat lost by iron piece
Q2=200 × 10-3 Kg × 470 J/kg-K (θ° C-80° C)
heat lost by iron piece=heat gained by calorimeter.
thus,
1255.8 J/K × 58° C=200 × 10-3 Kg × 470 J/kg-K (θ° C-80° C)
calculating for θ we get
θ=858.4° C

Question 15
1 kg of ice at 0 °C is mixed with 1 kg of steam at 100 °C. What will be the composition of the system when thermal equilibrium is reached. latent heat of fusion of ice = 3.36X105 J/kg and latent heat of vaporization of water = 2.26X106 J/kg.

Heat required to melt ice at 0 °C to water = 1 kg × 3.36 × 105 J / K.
= 336000 J
heat required to take 1 kg of water to temperature 100 from 0 °C to convert into water
=4200 × 100J
Total Heat(Q1)=336000+420000=756000=7.56X105
heat lost by steam at 100°C to convert into water
Q2=2.26 × 106 J/kg.
=22.6 X105
since Q2 > Q1. It means whole steam will not be cooled down .Some steam will cooled down to make ice reach the temperature of 100°C.it means thermal Equilibrium is reached at 100 °C and some mass of steam will be cooled down
Let m be the mass of steam cooled down
Heat lost=mX22.6 X105
From principal of calorimetry heat lost = heat gained
7.56X105 = mX22.6 X105
or m = .334
= >.334 kg of heat gets converted into water this total amount of water is 1.335 kg and steam left is .665 kg.

Question 16
A bullet of mass 30 gm. enters into a fixed wooden block with a speed 50 m / s and stops in it. Calculate the change in internal energy during this process.

given that velocity of bullet v = 50 m/s
and its mass is m = 30 gm
internal energy of the system is Kinetic energy plus Potential Energy of the system. Initially all the energy of system is kinetic which become potential when bullet enters wooden block so change in internal energy
Δu = (1 / 2 )mu2
= 1 / 2 (.30) × (50)2
= 37.5J

Question 17
An animal of 70 kg is running with a speed of 6 m/s. If all the Kinetic energy of animal can be used in increasing water from 18 °C to 32 °C, how much water can be heated with this energy

KE of animal = 1/2 × 70 × (6)2
= 12.60J
Q = 1260 J is the amount of heat being given to heat the water
since, Q = ms Δθ
m = Q/ sΔθ
= 12605 / 4186 × (32-18)
= 0.0215 kg = 21.5 gm

Question 18
A copper cube of mass 300g slides down on a rough inclined plane of inclination 40 ° at a constant speed. Assume that any loss in Mechanical energy goes into copper block as thermal energy find the increase in thermal energy of the block as it slides down through 50 cm. specific heat capacity of copper = 420J/kg-k. ?

work done in sliding 50 cm=Change in Potential energy=mgh
=mg(dsin40)
W = = .95 J.
If this mechanical work W produces same temperature change as heat Q then
W = J Q
where J = mechanical equivalent of heat and J = 1 if same unit of W and Q are used
Q= .95 J
since, Q= msΔθ
change in temperature Δθ of block
Δθ= Q/ ms
= .95 / .3 x 420
= 7.53 x 10-3 °C.