Chapter 21: Problem 2
Absolute zero is \(-273.15^{\circ} \mathrm{C}\). Find absolute zero on the Fahrenheit scale.
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Chapter 21: Problem 2
Absolute zero is \(-273.15^{\circ} \mathrm{C}\). Find absolute zero on the Fahrenheit scale.
These are the key concepts you need to understand to accurately answer the question.
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A rod is measured to be \(20.05 \mathrm{~cm}\) long using a steel ruler at a room temperature of \(20^{\circ} \mathrm{C}\). Both the rod and the ruler are placed in an oven at \(270^{\circ} \mathrm{C}\), where the rod now measures \(20.11 \mathrm{~cm}\) using the same rule. Calculate the coefficient of thermal expansion for the material of which the rod is made.
A thermocouple is formed from two different metals, joined at two points in such a way that a small voltage is produced when the two junctions are at different temperatures. In a particular iron-constantan thermocouple, with one junction held at \(0^{\circ} \mathrm{C}\), the output voltage varies linearly from 0 to \(28.0 \mathrm{mV}\) as the temperature of the other junction is raised from 0 to \(510^{\circ} \mathrm{C}\). Find the temperature of the variable junction when the thermocouple output is \(10.2 \mathrm{mV}\).
At what temperature is the Fahrenheit scale reading equal to (a) twice that of the Celsius and \((b)\) half that of the Celsius?
Repeat Exercise 1, except choose the new temperature scale \(\mathrm{Q}\) so that absolute zero is \(0^{\circ} \mathrm{Q}\) and \(T_{\mathrm{bp} \text { , water }}-T_{\text {mp, water }}=\) \(100 \mathrm{Q}^{\circ} .(a)\) What is the conversion formula from Celsius to Q? (b) What is \(T_{\text {bp, water }}\) and \(T_{\text {mp,water }}\) in Q? ( \(c\) ) This scale actually exists. What is the official name?
At \(100^{\circ} \mathrm{C}\) a glass flask is completely filled by \(891 \mathrm{~g}\) of mercury. What mass of mercury is needed to fill the flask at \(-35^{\circ} \mathrm{C} ?\) (The coefficient of linear expansion of glass is \(9.0 \times 10^{-6} / \mathrm{C}^{\circ} ;\) the coefficient of volume expansion of mercury is \(1.8 \times 10^{-4} / \mathrm{C}^{\circ}\).)
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