Wednesday, October 30, 2019

H C Verma solutions, Heat and Temperature, Exercises, Q1_Q10, Chapter-23, Concepts of Physics, Part-II

Heat and Temperature

Exercises - Q1 to Q10



1. The steam point and the ice point of a mercury thermometer are marked as 80° and 20°. What will be the temperature in centigrade mercury scale when this thermometer reads 32°?  

 


Answer: Here tᵢ = 20°, tₛ = 80°, t = 32°. Hence the temperature in centigrade corresponding to t = 32° will be
T = (t - tᵢ)*100/(tₛ - tᵢ)
   = (32° - 20°)*100/(80° - 20°)
   = 12°*100/60°
   = 20 °C  

It can also be solved by establishing a relationship between these two scales. Let us call the scale of the marking as S. So the 100 °C corresponds to 80 °S and 0 °C to 20 °S. Let the relationship between them be
C = aS + b
where a and b are constants.
Putting both the values we have two equations,
100 = 80a + b --------- (i)
0 = 20a + b ------------(ii)

Subtracting we get
100 = 60a, →a = 100/60 =5/3
Adding we get
100 = 100a + 2b
→2b =100 - 100*5/3 = -200/3
→b = -100/3
So the relationship is now
C = 5S/3 - 100/3 =(5S - 100)/3
→3C = 5S - 100
Now for the 32 °S,
3C = 5*32 -100 =160 - 100 = 60
→C = 60/3 = 20 °
So the temperature in centigrade mercury scale will be 20°C.




2. A constant volume thermometer registers a pressure of 1.500x10⁴ Pa at the triple point of water and pressure of 2.050x10⁴ Pa at the normal boiling point. What is the temperature at the normal boiling point? 

 

Answer: We know that for a constant volume gas thermometer, the temperature is given as
T = (p/pₜᵣ)*273.16 K
(Where T is the temperature corresponding to the pressure p)
Given pₜᵣ = 1.500x10⁴ Pa
p = 2.050x10⁴ Pa
Hence the temperature at the normal boiling point
T =(2.050x10⁴/1.500x10⁴)*273.16 K
   = 373.3 K



3. A gas thermometer measures the temperature from the variation of pressure of a sample of gas. If the pressure measured at the melting point of lead is 2.20 times the pressure measured at the triple point of water, find the melting point of lead.  

 

Answer: Here, p=2.20*pₜᵣ 
The temperature corresponding to p is 
T = (p/pₜᵣ)*273.16 K
   = 2.20*273.16 K
   = 601 K.


4. The pressure measured by a constant volume gas thermometer is 40 kPa at the triple point of water. What will be the pressure measured at the boiling point of water (100 °C)?  

 

Answer: pₜᵣ = 40 kPa
The boiling point of water T =100°C = 373.16 K and corresponding pressure =p. Hence
T = (p/pₜᵣ)*273.16 K
→373.16 = (p/40)*273.16 K
p = 40*373.16/273.16 kPa
   = 55 kPa.


5. The pressure of the gas in a constant volume gas thermometer is 70 kPa at the ice point. Find the pressure at the steam point.  

 

Answer: Pressure at ice point, pᵢ = 70 kPa. The temperature at ice point
T = (pᵢ/pₜᵣ)*273.16 K 
But T = 273.15 K
→273.15 = (70/pₜᵣ)*273.16 K ---(i)
The temperature T' at steam point corresponding to the pressure pₛ is
T' = (pₛ/pₜᵣ)*273.16 K
But T' = 373.15 K
→373.15 = (pₛ/pₜᵣ)*273.16 K ---(ii)
Dividing (ii) by (i) we get
373.15/273.15 = (pₛ/70)
→pₛ = 70*373.15/273.15 kPa
      = 96 kPa.


6. The pressures of the gas in a 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. 

 

Answer: Pressure at ice point, pᵢ =80 cm
The pressure at the steam point, pₛ = 90 cm
Pressure for heated wax bath, p₀ = 100 cm of mercury.
The temperature at ice point, tᵢ = 0 °C
The temperature at the steam point, tₛ = 100 °C
The temperature of the heated wax bath, t₀ =?.
We know, the temperature
t₀ = {(p₀ - pᵢ)/(pₛ - pᵢ)}*100 °C
    = {(100-80)/(90-80)}*100 °C
    = {20/10}*100 °C
    = 200 °C.


7. In a Callender's compensated constant pressure air thermometer, the volume of the bulb is 1800 cc. When the bulb is kept immersed in a vessel, 200 cc of mercury has to be poured out. Calculate the temperature of the vessel.  

 

Answer: The volume of the bulb, V =1800 cc.
The volume of the mercury taken out, v' = 200 cc. 
So the temperature of the vessel,
T = {V/(V-v')}*T₀, where T₀= Ice bath temperature = 273.15 K.
→T = {1800/(1800-200)}*273.15 K
      ={1800/1600}*273.15 K
      = 307 K.


8. A platinum resistance thermometer reads 0° when its resistance is 80 Ω and 100° when its resistance is 90 Ω. Find the temperature at the platinum scale at which the resistance is 86 Ω.

 

Answer: Given Rₜ = 86 Ω, R₀ = 80 Ω, R₁₀₀ = 90 Ω. The temperature corresponding to Rₜ is given as,
t = {(Rₜ - R₀)/(R₁₀₀ - R₀)}*100°
  = {(86-80)/(90-80)}*100°
  = {6/10}*100°
  = 60°.


9. A resistance thermometer reads R = 20.0 Ω, 27.5 Ω and 50 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+ßθ²), find the values of R₀, α, and ß. Here θ represents the temperature on the Celsius scale.


Answer: With the given resistance-temperature relation three equations can be formed from which three unknowns can be found out. For θ = 0°C, Rθ = 20.0 Ω, putting in the given relation
20.0 Ω= R₀(1+α*0+ß*0) =R₀
→R₀ = 20.0 Ω
For θ = 100 °C, Rθ = 27.5 Ω, hence
27.5 = 20.0(1+α*100+ß*10000)
→100α+10000ß =(27.5/20)-1=1.375-1
→α+100ß = 0.375/100
→α+100ß =3.75x10⁻³ ------------(i)
Also for θ = 420 °C, Rθ = 50.0 Ω, So
50.0 = 20.0(1+420α+420²ß)
→α + 420ß ={(50.0/20.0)-1}/420
→α + 420ß = 1.5/420=3.57x10⁻³ ---(ii)
Subtracting (i) from (ii)
320ß =(3.57-3.75)x10⁻³ =-(0.18)*10⁻³
→ß =-(0.18/320)*10⁻³
→ß =-(0.00056)*10⁻³ 
→ß =-5.6x10⁻⁷/°C²
From (i),
α = 3.75x10⁻³ - 100ß
→α =3.75x10⁻³ - 100(-5.6x10⁻⁷)
→α = 3.72x10⁻³ + 0.056x10⁻³
→α = 3.776x10⁻³ °C
→α ≈ 3.8x10⁻³ °C

   

10. A concrete slab has a length of 10 m on a winter night when the temperature is 0 °C. Find the length of the slab on a summer day when the temperature is 35 °C. The coefficient of linear expansion of concrete is 1.0x10⁻⁵/°C. 

 

Answer: Given, L₀ = 10 m
α = 1.0x10⁻⁵/°C, θ = 35 °C.
The length at 35 °C is given as
Lθ = L₀(1+αθ)
    = 10(1+1.0x10⁻⁵*35) m
    = 10(1.00035) m
    = 10.0035 m 

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CHAPTER- 20 - Dispersion and Spectra


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CHAPTER- 14 - Fluid Mechanics



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CHAPTER- 3 - Kinematics - Rest and Motion

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