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1.

A platinum resistancethermometer reads `0^0`at the ice point and boiling point of water respectively. The resistance of a platnum wire varies with Celsius temperature `theta` as `R_t = R_0 (1+alphatheta + betatheta^2)`, where `alpha = 3.8 xx 10^(-3) ^0 C^(-1)` and `beta= -5.6 xx 10^(-7) ^0 C^(-2)`. What will be the reading of this thermometer if it is pleaced in a liquid bath maintained at `50^0 C`?

Answer» The resistances of the wire in the thermometer at `100^0 C` and `50^0 C` are
`R_100 = R_0 [1 + alpha xx 100^0 C + beta xx (100^0 C)^2 ]
and, ` R_50 = R_0 [ 1 + alpha xx 50^0 C + beta xx (50^0 C)^2]`.
The temperature `t` measured on the platinum thermometer is given by
`t = (R_50 - R_0)/(R_100 - R_0) xx 100^0`
= (alpha xx 50^0 C + beta xx (50^0 C)^2)/(alpha xx 100^0 C + beta xx (100^0 C)^2 ) xx 100^0`
`=50.4^0`.
Discussion
2.

The pressure of the gas in a constant volume gas thermometer is 80cm of mercury in melting ice at `1 atm`. When the bulb is placed in a liquid, the pressure becomes `160cm` of mercury. Find the temperature of the liquid.

Answer» For an ideal gas at constant volume,
`T_1/T_3 = P_1/P_3`
or, T_2 = P_2/P_1 T_1`. The temperature of melting ice at `1 atm` is `273.15K`. Thus, the temperature of the liquid is
`T_2 = 160/80 xx 273.15 K = 546.30K`.
Discussion
3.

In a constant volume gas thermometer, the pressure of the working gas is measured by the differenced in the levels of mercury in the two arms of a U-tube connected to the gas at one end. When the bulb is placed at the room temperature `27.0^0 C`, the mercury column in the arm open to atmosphere stands `5.00 cm` above the level of mercury in the other arm. When the bulb is placed in a hot liquid, the difference of mercury levels becomes `45..0 Cm`. Calculate the temperature of the liquid. (Atmospheric pressure `= 75.0cm` of mercury).

Answer» The pressure of the gas = atmospheric pressure + the pressure due to the difference in mercury levels At `27^0 C`, the pressure is `75cm + 5cm = 80cm` of mercury. At the liquid temperature, the pressure is `75cm + 45cm = 120cm` of mercury. Using `T_2 = P_2/P_1 T_1`, the temperature of the liquid is
`T= 120/80 xx (27.0 + 273.15) K = 450.22K`.
`= 177.07^0 C = 177^0 C`.
Discussion
4.

A resistance thermometer reads `R=20.0Omega, 27.5 Omega, and 50.0 Omega` at the point `(0^C)`, the steam point `(100^C)` and the zinc point `(420^C)` respectively.Assuming that the resistance varies with temperature as `R_Theta=(1+alphaTheta+betaTheta^2)` , find the values of `R_0`, alpha and beta. Here Theta represents the temperature on Celsius scale.

Answer» `R at ice point (R_0) =20Omega`
`R` at steam point `(R_100)=27.5Omega`
`R at zinc point (R_420)=50 Omega`
`R_theta = R _0 (1 + alpha theta+beta theta^2)`
`rArr R_100 =R_0+R_0 alpha theta+ R_0 beta theta^2`
`rArr (R_100-R_0)/R_0 = alpha theta +beta theta^2`
`rArr 27.5-20/20=alpha theta+ beta theta^2`
`rArr 7.5/20= alpha100+beta 10000 ........(i)`
Again `R_420= R_0(1+alpha theta+beta theta^2)`
`rArr 50-R_0/20 = alpha theta+ beta theta^2`
`50-20/20 = 420alpha +176400 beta`
`rArr 3/2=420 alpha+176400 beta .............(ii)`
After solving (i) and (ii), we get
`alpha =3.8 xx 10^(-3)/^0C`,
`beta = -5.6 xx 10^(-7)/^0C`
Discussion
5.

A concrete slab has a length of 10 m on a winter night when the temperature is `0^C`. Find the lenth of the slab on a summer day when the temperature is `35^C`. The coefficient of linear expansion of concrete is `1.0 XX 10^(-5) 0^ C^(-1)`.

Answer» Given `L_1=? L_0= 10m`
`alpha =1 xx 10^(-5)/^0C`
`t=35^0C`
`L_1= L_0( 1+alpha t)`
` =10(1+10^(-5)xx 35)`
`10+35xx10^-4`
`10.0035m`
Discussion
6.

The pressue of air in the bulb of a constant volume gas thermometer of `0^0 C` and `100^0 C` are `73.00cm and 100cm` of mercury respectively.Calculate the pressure at the room temperature `20^0 C`.

Answer» `t= (P_1 - P_0)/P_100 - P_0) xx 10^0 C`.
Thus `20^0 C= (P_i - 73.00 cm of Hg)/(100cm of Hg - 73.00cm of Hg) xx 10^0 C`
or, `P_t = 78.4cm` of mercury.
Discussion