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Question: On a hypothetical scale A the ice point is \({{42}^{{}^\circ }}\) and the steam point is \({{182}^{{...

On a hypothetical scale A the ice point is 42{{42}^{{}^\circ }} and the steam point is 182{{182}^{{}^\circ }}. For another scale B, the ice point is 10-{{10}^{{}^\circ }} and the steam point is 90{{90}^{{}^\circ }}. If B reads 60{{60}^{{}^\circ }}, what is the reading of A?
A. 160{{160}^{{}^\circ }}
B. 140{{140}^{{}^\circ }}
C. 120{{120}^{{}^\circ }}
D. 110{{110}^{{}^\circ }}

Explanation

Solution

Assume some arbitrary linear relation between scale A and scale B. Use the given data to obtain equations with the assumed relation. Determine the constants value in assumed relation. Now, the obtained relation between scale A and B can be used to determine convert temperature from one scale to another.

Complete step by step answer:
Let us assume that temperature scale A and temperature scale B are related by the following relation.
pTB+q=TAp{{T}_{B}}+q={{T}_{A}}
Where TA{{T}_{A}} and TB{{T}_{B}} are the temperatures as measured by temperature scale A and B respectively and pp and qq are arbitrary constants.
For hypothetical scale A ice point temperature is 42{{42}^{{}^\circ }} and for scale B ice point is at temperature 10-{{10}^{{}^\circ }}. Substituting these values in above relation, we get
10p+q=42-{{10}^{{}^\circ }}p+q={{42}^{{}^\circ }} ….. (1)
Similarly, for steam point, temperature measured by scale A and B are 182{{182}^{{}^\circ }} and 90{{90}^{{}^\circ }} respectively. Substituting these values, we get
90p+q=182{{90}^{{}^\circ }}p+q={{182}^{{}^\circ }} ….. (2)
Using elimination method for solving equation (1) and (2), we have
90p+q(10p+q)=18242{{90}^{{}^\circ }}p+q-\left( -{{10}^{{}^\circ }}p+q \right)={{182}^{{}^\circ }}-{{42}^{{}^\circ }}
This implies that,
100p=140100p=140
Or p=75p=\dfrac{7}{5}
Substituting this value in either of the equation, we get q=56q=56
The relation between temperature scale A and B becomes
TA=75TB+56{{T}_{A}}=\dfrac{7}{5}{{T}_{B}}+56
If temperature noted by hypothetical scale B is 60{{60}^{{}^\circ }}, then temperature observed by scale A is
TA=75×60+56{{T}_{A}}=\dfrac{7}{5}\times 60+56
TA=140{{T}_{A}}={{140}^{{}^\circ }}

So, the correct answer is “Option B”.

Note: The ice point and steam point are very sensitive to pressure, presence of impurities in liquid and nature of the container used to hold the liquid. Therefore, for scale calibration, the triple point of water is assigned a fixed value of temperature and other temperatures are measured with reference to it.