## Measuring temperature

### How do we measure temperature?

Assume we have a gas in a container with a movable piston under atmospheric pressure.  As we increase the temperature of the gas it expands, its volume increases.  The piston moves out.  As we cool the gas the piston moves in.  The ideal gas law gives us the volume of the gas as a function of temperature.

PV = NkBT,    V = (NkB/P)T.

The volume is proportional to the absolute temperature of the gas.  Assume the temperature change from 0 oC to 10oC.  We have

(ΔV/V) = (ΔT/T) = 10/273 = 0.037 = 3.7%.

By monitoring the volume of the gas we can monitor its temperature.  We have a gas thermometer.

Solids and liquids also expand as the temperature increases.  For most solids we can define an average coefficient of linear expansion, α.  The change in length Δl of a solid is proportional to its length l and the change in temperature ΔT.  The proportional constant is α.

Δl = αlΔT.

The average volume expansion coefficient β is defined through ΔV = βVΔT.  We have β = 3α.

Linear expansion coefficients α: (per oC)

Aluminum 23 * 10-6 17 * 10-6 9 * 10-6 3.2 * 10-6 11 * 10-6 0.7 * 10-6

For some applications it is important to choose materials that have a small coefficient of linear expansion.  Temperature variations can destroy a precision alignment.  For other applications it is important that all materials have similar coefficients of linear expansion.  When different parts of a structure expand by different amounts, the structure buckles.

#### Problem:

How much taller does the Eiffel Tower become at the end of a day when the temperature has increased by 15 oC?  Its original height is 321 m and you can assume it is made of steel.

Solution:
Δl = αlΔT = (11*10-6/oC)*(321 m)*(15 oC) = 5.3*10-2 m = 5.3 cm.
The height of the tower increases by 5.3 cm.

#### Problem:

A copper telephone wire has essentially no sag between poles 35 m apart on a winter day when the temperature is -20 oC.  How much longer is the wire on a summer day, when the temperature is 35 oC?

Solution:
Δl = αlΔT.  For copper α = 17*10-6 per oC.  Δl = (17*10-6)(35 m)55 = 0.0327 m = 3.27 cm.

#### Problem:

A square hole 8 cm along each side is cut in a sheet of copper.  Calculate the change in the area of the hole if the temperature of the sheet is increased by 50 K.

Solution:
ΔA = 2αAΔT.  For copper α = 17*10-6 per oC.  ΔA = (34*10-6)(64 cm2)50 = 0.109 cm2.

#### Problem:

How large an expansion gap should be left between steel railroad rails if they may reach a maximum temperature 35.0 oC greater than when they were laid?  Their original length is 10.0 m.

Solution:
Δl = αlΔT = (11*10-6/oC)*(10 m)*(35 oC) = 3.85*10-3 m ~ 4 mm.

If we want to build a thermometer, we want to use a material with a large coefficient of volume expansion.  For most liquids, the coefficient of volume expansion is on the order of 10-4.  For each degree Celsius temperature change, the volume changes by (ΔV/V) = β = 10-4 = 0.0001 = 0.01%.  If we tried to observe this change by monitoring the height of water in a glass, we would probably be unsuccessful.

Assume you have 1 liter = 1000 cm3 of liquid with β = 10-4 in a container with bottom area A = 100 cm2
The height of the liquid is 1 cm.  (Volume = area * height.)
You increase the temperature of the liquid by 20 oC.
ΔV = βVΔT = 10-4 * 1000 cm3 * 20 = 2 cm3
The change in height is Δh = ΔV/A = 2 cm3/(100 cm2) = 0.02 cm = 0.2 mm.

If, however, you put a tight lid on the container at a height of 10 cm with a hole connected to a fine capillary, then the liquid will rise in the capillary to a much greater height.  Assume you connected a tube with a cross sectional area of 1 cm2.  Then Δh = ΔV/A = 2 cm3/(1 cm2) = 2 cm.  You have constructed a usable thermometer.

Thermometers can also be constructed from bimetallic strips.  Two strips of metal, one of steel and one of aluminum, that have the same length at 0 oC, will have different length a 200 oC.  Assume the strips are 10 cm long at 0 oC.  The change in length of the steel strip at 200 oC is Δl = αlΔT = 11*10-6*10 cm*200 = 0.022 cm = 0.22 mm and the change in length of the aluminum strip at 200 oC is Δl = αlΔT = 23*10-6*10 cm*200 = 0.046 cm = 0.46 mm.  If the two strips are bonded together, the bimetallic strip will buckle or curl.  Long bimetallic coils will uncoil slightly as the temperature rises if the inner strip has the larger α.  If one end is fixed and a pointer is attached to the other end the pointer will move.  The position of the pointer indicates the temperature.