A mercury thermometer has a bulb of volume 0.100 cm^{3} at 10^{o}C. The
capillary tube above the bulb has a cross-sectional area of 0.012 mm^{2}.
The volume thermal expansion coefficient of mercury is 1.8*10^{-4} (^{o}C)^{-1}.
How much will the mercury rise when the temperature rises by 20^{o}C?

Solution:

- Concepts:

The volume expansion coefficient

The average volume expansion coefficient b is defined through ΔV = βVΔT. - Reasoning:

The volume of the mercury will increase as the temperature rises and the mercury will rise in the capillary tube. - Details of the calculation:

The increase in temperature is 20^{o}C. The change in the volume of the mercury is

ΔV = βVΔT = (1.8*10^{-4}(^{o}C)^{-1})(0.100 cm^{3})(20^{o}C) = 3.6*10^{-4}cm^{3}= 0.36 mm^{3}.

Δh = ΔV/A = (0.36 mm^{3})/(0.012 mm^{2}) = 30 mm.

The mercury will rise 30 mm.

A bar of gold is in thermal contact with a bar of silver of the same length
and area. One end of the compound bar is maintained at 80 ^{o}C and the
opposite end is at 30 ^{o}C. When the heat flow reaches steady state,
find the temperature at the junction.

The thermal conductivity of gold is 314 W/(m^{o}C), and the thermal
conductivity of silver is 427 W/(m^{o}C).

Solution:

- Concepts:

Thermal conductivity

The thermal conductivity k is defined through the equation ΔQ/Δt = -kA ΔT/Δx. - Reasoning:

When a**steady state**is reached, then**the same amount of heat crosses any cross sectional area perpendicular to the bar per second**. (Otherwise the energy would increase in certain regions, the temperature would increase there, the temperature gradient would change, and we would not have a steady state.) The amount of heat flowing is determined by the temperature difference and the thermal conductivity. - Details of the calculation:

If the temperature at the junction is T, then in the gold we have

ΔQ/Δt = (314 W/(m^{o}C))*A*(80^{o}C - T)/(L/2)

and in the silver we have

ΔQ/Δt = (427 W/(m^{o}C))*A*(T - 30^{o}C)/(L/2).

We can therefore write

(314 W/(m^{o}C))*A*(80^{o}C - T)/(L/2) = (427 W/(m^{o}C))*A*(T - 30^{o}C)/(L/2).

(314 W/(m^{o}C))*(80^{o}C - T) = (427W/(m^{o}C))*(T - 30^{o}C).

(80^{o}C - T) = 1.36*(T - 30^{o}C).

120.8^{o}C = 2.36 T.

T = 51.2^{o}C.

A Thermopane window of area 6 m^{2} is constructed of two layers of
glass, each 4 mm thick, separated by an air space of 5 mm. If the inside is at
20 ^{o}C and the outside is at -30 ^{o}C, what is the rate of
heat loss through the window?

Thermal conductivity of Thermopane: 0.8 W/(m^{o}C)

Thermal conductivity of air: 0.0234 W/(m^{o}C)

Solution:

- Concepts:

Thermal conductivity

The thermal conductivity k is defined through the equation. - Reasoning:

When a steady state is reached, then the same amount of heat crosses any cross sectional area perpendicular to the bar per second. (Otherwise the energy would increase in certain regions, the temperature would increase there, the temperature gradient would change, and we would not have a steady state.) The amount of heat flowing is determined by the temperature difference and the thermal conductivity.

ΔQ/Δt = -kA ΔT/Δx. - Details of the calculations

When a steady state is reached, then the same amount of heat crosses any cross sectional area per second. Let the temperature of the inner glass-air boundary be T_{1}and the temperature of the outer glass-air boundary be T_{2}. Then for the inner piece of glass we have

ΔQ/Δt = (0.8 W/(m^{o}C))*6 m^{2}*(20^{o}C - T_{1})/0.004 m.

For the air layer we have

ΔQ/Δt = (0.0234 W/(m^{o}C))*6 m^{2}*(T_{1 }- T_{2})/0.005 m.

For the outer piece of glass we have

ΔQ/Δt = (0.8 W/(m^{o}C))*6 m^{2}*(T_{2 }+ 30^{o}C)/0.004 m.

We have 3 equations for three unknowns, ΔQ/Δt, T_{1}, and T_{2}.

The first and third equation yield (20^{o}C- T_{1}) = (T_{2 }+ 30^{o}C), T_{2 }= -10^{o}C - T_{1}.

Inserting this expression for T_{2}into the second equation we have

ΔQ/Δt = (0.0234 W/(m^{o}C))*6 m^{2}*(2 T_{1}+10^{ o}C)/0.005 m.

Combining this equation with the first equation then yields

(0.0234 W/(m^{o}C))*(2 T_{1}+10^{o}C)/0.005 m = (0.8 W/(m^{o}C))*(20^{o }C - T_{1})/0.004 m.

0.0234*(2 T_{1}+10^{o}C) = 20^{o}C - T_{1}.

1.0468 T_{1 }= 19.776^{o}C. T_{1 }= 18.882^{o}C.

The first equation now yields

ΔQ/Δt = (0.8 W/(m^{o}C))*6 m^{2}*(1.118^{o}C)/0.004 m = 1.34 kW.