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Chapter 1: Problem 45

You decide to establish a new temperature scale on which the melting point of mercury \(\left(-38.9^{\circ} \mathrm{C}\right)\) is \(0^{\circ} \mathrm{M},\) and the boiling point of mercury \(\left(356.9^{\circ} \mathrm{C}\right)\) is \(100^{\circ} \mathrm{M} .\) What would be (a) the boiling point of water in \(^{\circ} \mathrm{M} ;\) and \((\mathrm{b})\) the temperature of absolute zero in \(^{\circ}\text{M}\)?

Short Answer

Expert verified
To answer the results are as follows: (a) The boiling point of water in M degrees is just the value you obtained from substituting T_C = 100 into the equation. (b) The absolute zero temperature in M degrees is the result you got from substituting T_C = -273.15 into the equation applied.

Step by step solution

01

Define the linear equation for the conversion

First, form a linear equation using the given mercury points. Since we know that at -38.9°C, the temperature is 0M, and at 356.9°C, it's 100M, the equation will be in the form: \(T_M = a T_C + b\). But for T_C = -38.9°C, T_M = 0M and for T_C = 356.9°C, T_M = 100M.
02

Calculation of the equation coefficients

To find the coefficients a and b, we'll substitute the given values into the equation. Starting with the known point (-38.9,0), we substitute these as T_C and T_M in the equation, obtaining: \(0 = a*(-38.9) + b\). And then, using the point (356.9,100): \(100=a*356.9 +b\). Solving these simultaneous equations gives us the values of a and b for the equation.
03

Calculating temperatures in Mercury scale

With the linear equation established, we can find the Mercury equivalent for any Celsius temperature. For example, to solve for part (a), where the boiling point of water is 100°C, substitute T_C = 100 into the equation to find T_M. For part (b), consider that absolute zero is -273.15°C in the Celsius scale, so substitute T_C = -273.15 to find T_M.

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