Problem 1.7 (Schroeder’s Intro to Thermal Physics)

Problem 1.7

When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, \beta:

\beta = \dfrac{\Delta V / V}{\Delta T}

(where V is volume, T is temperature, and \Delta signifies a change, which in this case should really be infinitesimal if \beta is to be well defined). So for mercury, \beta = 1/5500 \text{K}^{-1} = 1.81 \times 10^{-4} \text{K}^{-1}. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)

(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.

(b) The thermal expansion coecient of water varies significantly with temperature: It is 7.5 \times 10^{-4} \text{K}^{-1} at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of 0.68 \times 10^{-4} \text{K}^{-1} at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coecient of water were always positive.

Solution:

Problem 1.7 (Schroeder's Intro to Thermal Physics) part 1 of 2
Problem 1.7 (Schroeder's Intro to Thermal Physics) part 2 of 2

Problem 1.7 Solution (Download) 1 of 2

Problem 1.7 Solution (Download) 2 of 2

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