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 CO_{2} = 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 C0_{2} = 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 CO_{2} = (9/5)(-56.6) + 32

= - 69.88 ° F

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

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

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 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

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 R

Given are the values of resistances,

at ice point 0°C, R_{0} = 20.0Ω

at steam point 100°C, R_{100} = 27.5Ω

at zinc point 420°C, R_{z} = 50.0Ω

Now resistance R_{θ} varies as

R_{θ} = R_{0} (1+αθ+βθ^{2})

For resistance at ice point

R_{ice}= R_{0} (1 +α°C +β(0°C)^{2})

R_{ice} = R_{0} = 20°C

For resistance at steam point

R_{steam}= R_{0} (1 +α(100°C) +β(100°C^{2})

27.5α / 20.0α = 1 + 100 (α+ 100β)

or,

α+100β= 3.75 × 10^{-3} .......(i)

for resistance at zinc point

R_{z} = R_{0} [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}

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

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 = l_{0}(1 +αΔθ)

Where,

l_{0} = 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^{-5}X 100°C)

= 2.0046 cm.

The density of water at 0°C is 0.998 gm/cm

Density is mass per unit volume given in the question is density of water at 0°C and 4°C

ρ_{0} = .998 gm/cm^{3} and ρ_{4}=1.00 gm/cm^{3} with increase in temperature volume of water changes using,

V = V_{0} (1 +βθ)

β is the coefficient of volume expansion.

Volume at 4°C equals

V_{4} = V_{0}(1 +β (4°C - 0°C))

or V_{4} = V_{0} (1 + 4β)

If m is the mass of water then density of water at 4°C is

m/V_{4} =m/V_{0} [(1 + 4β)]

or ρ_{4} = ρ_{0} / (1 +4β)

1.00 =.998/(1+4β)

1 +4β = .998

β = -5 X 10^{-4} °C^{-1}

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)

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 cm

At temperature θ= 0°C Volume of glass vessel is

V = 1000 cm^{3}

It is then completely filled with mercury.

At temperature θ= 10°C, 1.6 cm^{3} 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}=V_{0} (1+γ_{m}Δθ)

and that of glass.

V_{gθ}=V_{0} (1+γ_{g}Δθ)

now volume of mercury overflown is

V_{mθ}-V_{gθ} = V_{0}(γ_{m}-γ_{g})Δθ

1.6 cm^{3} = 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}.

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

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}

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 CO

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

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 α

(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,

D_{θ }= D_{20 }(1+ α_{S}Δθ)

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.

Specific heat capacity of the a metal A is expressed

as

C=K

How much heat energy is required to raise the temperature from T

a. (T

b. (T

c. (T

d. None of the above

dQ=CdT

Integrating with limits T_{0} from 2T_{0}

∫dQ=∫(K_{1}T+K_{2}T^{3})dT

Q=(T_{0}^{2}/4)(6K_{1} +15K_{2}T_{0}^{2})

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

Q_{1}=.35*4200(T-18)

Heat gained by the aluminium

Q_{2}=.65*910(T-18)

Heat lost by iron block

Q_{3}=.2*4700(100-T)

Now

Q_{1}+Q_{2}=Q_{3}

Substituting all values and solving

T=21.58° C

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

Q_{1}=300 × 10^{-3} Kg × 4186 J/kg-K (80° C - 22° C)

or Q_{1}=1255.8 J/K × 58° C

now, heat lost by iron piece

Q_{2}=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

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 = 3.36X10

Heat required to melt ice at 0 °C to water = 1 kg × 3.36 × 10^{5} 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(Q_{1})=336000+420000=756000=7.56X10^{5}

heat lost by steam at 100°C to convert into water

Q_{2}=2.26 × 10^{6} J/kg.

=22.6 X10^{5}

since Q_{2} > Q_{1}. 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 X10^{5}

From principal of calorimetry heat lost = heat gained

7.56X10^{5} = mX22.6 X10^{5}

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.

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 )mu^{2}

= 1 / 2 (.30) × (50)^{2}

= 37.5J

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

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.

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