Solved Problem on Temperature
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The temperature can be measured with a mercury thermometer. In this type of thermometer, the temperature is the length ℓ of a capillary column, measured from a common origin. It is verified that ℓ = 2.34 cm, when the thermometer is in thermal equilibrium with melting ice, and ℓ = 12.34 cm when thermal equilibrium is with boiling water (in an environment at which atmospheric pressure is 1 atm).
a) Determine the equation of this thermometer;
b) Calculate the length of the column of mercury when the temperature is θ = 25 °C;
c) Calculate the temperature of the environment when ℓ = 8.84 cm.


Problem data:
  • Length of the capillary when the thermometer is in thermal equilibrium with the melting ice:    ℓ1 = 2.34 cm;
  • Length of the capillary when the thermometer is in thermal equilibrium with boiling water:    ℓ2 = 12.34 cm.
Problem diagram:

eFigura 1

The problem does not provide us with temperatures of melting ice and boiling water, but tells us that the environment is the atmospheric pressure of 1 atm, so
  • Temperature of ice fusion:    θ1 = 0 °C;
  • Temperature of boiling water:    θ2 = 100 °C.
Solution

a) To find the equation we wrote
\[ \bbox[#99CCFF,10px] {\frac{g-g_{1}}{g_{2}-g_{1}}=\frac{t-t_{1}}{t_{2}-t_{1}}} \]
where temperature g for this problem is the length of the column of mercury ℓ
\[ \frac{\ell-\ell_{1}}{\ell_{2}-\ell_{1}}=\frac{\theta -\theta _{1}}{\theta_{2}-\theta _{1}} \]
substituting the values ​​given in the problem, we have
\[ \begin{gather} \frac{\ell-2.34}{12.34-2.34}=\frac{\theta-0}{100-0}\\ \frac{\ell-2.34}{10}=\frac{\theta }{100}\\ \theta=10\cancel{0}\times \frac{\ell-2.34}{\cancel{10}} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {\theta =10(\ell-2.34)} \]

b) To find the length of the mercury column, we will use the expression found above, for θ = 25 ºC we, have
\[ \begin{gather} 25=10(\ell-2.34)\\ 25=10\ell-10\times 2.34\\ 10\ell=25+23.4\\ \ell=\frac{48.4}{10} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {\ell=4.84\;\text{cm}} \]

c) For ℓ = 8.84 cm, using item expression (a), we have
\[ \begin{gather} \theta =10(8.84-2.34)\\ \theta =10\times 8.84-10\times 2.34\\ \theta=88.4-23.4 \end{gather} \]
\[ \bbox[#FFCCCC,10px] {\theta =65\;\text{°C}} \]
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