TITLE.PM5

50 ENGINEERING THERMODYNAMICS

dharm
M-therm/th2-2.pm5

(ii)Mean specific heat of the gas, cn :

Q = m.cn.dT

i.e., 420 = 6 × cn × (125 – 25)

∴ cn =

420

6 100× = 0.7 kJ/kg°-C. (Ans.)

Example 2.12. A temperature scale of certain thermometer is given by the relation

t = a ln p + b

where a and b are constants and p is the thermometric property of the fluid in the thermometer.

If at the ice point and steam point the thermometric properties are found to be 1.5 and 7.5

respectively what will be the temperature corresponding to the thermometric property of 3.5 on

Celsius scale. (Poona University, Nov. 2001)

Solution. t = a ln p + b ...(Given)

On Celsius scale :

Ice point = 0°C, and

Steam point = 100°C

∴From given conditions, we have

0 = a ln 1.5 + b ...(i)

and 100 = a ln 7.5 + b ...(ii)

i.e., 0 = a × 0.4054 + b ...(iii)

and 100 = a × 2.015 + b ...(iv)

Subtracting (iii) from (iv), we get

100 = 1.61a

or a = 62.112

Substituting this value in eqn. (iii), we get

b = – 0.4054 × 62.112= – 25.18

∴ When p = 3.5 the value of temperature is given by

t = 62.112 ln (3.5) – 25.18= 52.63°C. (Ans.)

Example 2.13. A thermocouple with test junction at t°C on gas thermometer scale and

reference junction at ice point gives the e.m.f. as

e = 0.20 t – 5 × 10 –4t^2 mV.

The millivoltmeter is calibrated at ice and steam points. What will be the reading on this

thermometer where the gas thermometer reads 70°C?

Solution. e = 0.20 t – 5 × 10–4t^2 mV ...(Given)

At ice point : When t = 0°C, e = 0

At steam point : When t = 100°C,

e = 0.20 × 100 – 5 × 10–4 × (100)^2 = 15 mV

Now, when t = 70°C

e = 0.20 × 70 – 5 × 10–4 × (70)^2 = 11.55 mV

∴ When the gas thermometer reads 70°C the thermocouple will read

t = 100 11.55

15

× = 77 °C. (Ans.)