thermal properties of matter important problems

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:-
T_C = T_K - 273
Where T_C is temperature on Celsius scale
T_K 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:-
T_F = (9T_C/ 5 )+ 32
T_F is temperature on Fahrenheit scale.
T_C 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

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 T_A on absolute scale A = (273.16XT_A)/200
Similarly value of temperature T_B on absolute scale B = (273.16XT_B)/350
Since T_A and T_B represent the same temperature
273.16×T_A/200 = 273.16×T_B/350
Or, T_A = 200T_B/350 = 4T_B/ 7

Given that,
Pressure at ice point P_ice = 80 cm Hg
Pressure at steam P_steam = 90 cm Hg
Pressure at heated wax bath P_wax = 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

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, R_z = 50.0Ω
Now resistance R_θ varies as
R_θ = R₀ (1+αθ+βθ²)
For resistance at ice point
R_ice= R₀ (1 +α°C +β(0°C)²)
R_ice = R₀ = 20°C
For resistance at steam point
R_steam= 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
R_z = 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

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.

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

(a) and (d)

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 γ_m
Coefficient of volume expansion of glass γ_g= 3α
γ_g = 3 × 6.5 × 10^-6 °C^-1
= 1.95 × 10^-5 °C^-1
Volume expansion of mercury at 10°C
V_mθ=V₀ (1+γ_mΔθ)
and that of glass.
V_gθ=V₀ (1+γ_gΔθ)
now volume of mercury overflown is
V_mθ-V_gθ = V₀(γ_m-γ_g)Δθ
1.6 cm³ = 1000 × 10°C (γ_m - 1.95 × 10^-5)
γ_m = 1.795 × 10^-4
or approximately
γ_m = 1.8 × 10^-4 °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

Given the diameter of hole = 0.2500 inch
at -78°C temperature of dry ice diameter of revert D_r = 0.2500 in
To find diameter of revert at 20°C using,
D_r20 = D_r (1 + α_al Δθ)
= 0.2500 (1 + 2.4 × 10^-5(20-(-78))
= .2506 inch

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,
D_θ = D₂₀ (1+ α_SΔθ)
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_rθ and D_sθ be the diameter of ring and shaft and are equal at temperature θ, then D_rθ= D_sθ=D
Using,
(i) for ring
D= 3[1+α_s(θ- 20)]
(ii) for shaft
D= 3.002[1+α_b(θ- 20)]
Subtracting (i) from (ii)
3.002+α_b × 3.002(θ- 20) - 3-3α_s(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.

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

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

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

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.

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

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

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.