
On a hypothetical scale A the ice point is ${{42}^{{}^\circ }}$ and the steam point is ${{182}^{{}^\circ }}$. For another scale B, the ice point is $-{{10}^{{}^\circ }}$ and the steam point is ${{90}^{{}^\circ }}$. If B reads ${{60}^{{}^\circ }}$, what is the reading of A?
A. ${{160}^{{}^\circ }}$
B. ${{140}^{{}^\circ }}$
C. ${{120}^{{}^\circ }}$
D. ${{110}^{{}^\circ }}$
Answer
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Hint: 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.
$p{{T}_{B}}+q={{T}_{A}}$
Where ${{T}_{A}}$ and ${{T}_{B}}$ are the temperatures as measured by temperature scale A and B respectively and $p$ and $q$ are arbitrary constants.
For hypothetical scale A ice point temperature is ${{42}^{{}^\circ }}$ and for scale B ice point is at temperature $-{{10}^{{}^\circ }}$. Substituting these values in above relation, we get
$-{{10}^{{}^\circ }}p+q={{42}^{{}^\circ }}$ ….. (1)
Similarly, for steam point, temperature measured by scale A and B are ${{182}^{{}^\circ }}$ and ${{90}^{{}^\circ }}$ respectively. Substituting these values, we get
${{90}^{{}^\circ }}p+q={{182}^{{}^\circ }}$ ….. (2)
Using elimination method for solving equation (1) and (2), we have
${{90}^{{}^\circ }}p+q-\left( -{{10}^{{}^\circ }}p+q \right)={{182}^{{}^\circ }}-{{42}^{{}^\circ }}$
This implies that,
$100p=140$
Or $p=\dfrac{7}{5}$
Substituting this value in either of the equation, we get $q=56$
The relation between temperature scale A and B becomes
${{T}_{A}}=\dfrac{7}{5}{{T}_{B}}+56$
If temperature noted by hypothetical scale B is ${{60}^{{}^\circ }}$, then temperature observed by scale A is
${{T}_{A}}=\dfrac{7}{5}\times 60+56$
${{T}_{A}}={{140}^{{}^\circ }}$
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.
Complete step by step answer:
Let us assume that temperature scale A and temperature scale B are related by the following relation.
$p{{T}_{B}}+q={{T}_{A}}$
Where ${{T}_{A}}$ and ${{T}_{B}}$ are the temperatures as measured by temperature scale A and B respectively and $p$ and $q$ are arbitrary constants.
For hypothetical scale A ice point temperature is ${{42}^{{}^\circ }}$ and for scale B ice point is at temperature $-{{10}^{{}^\circ }}$. Substituting these values in above relation, we get
$-{{10}^{{}^\circ }}p+q={{42}^{{}^\circ }}$ ….. (1)
Similarly, for steam point, temperature measured by scale A and B are ${{182}^{{}^\circ }}$ and ${{90}^{{}^\circ }}$ respectively. Substituting these values, we get
${{90}^{{}^\circ }}p+q={{182}^{{}^\circ }}$ ….. (2)
Using elimination method for solving equation (1) and (2), we have
${{90}^{{}^\circ }}p+q-\left( -{{10}^{{}^\circ }}p+q \right)={{182}^{{}^\circ }}-{{42}^{{}^\circ }}$
This implies that,
$100p=140$
Or $p=\dfrac{7}{5}$
Substituting this value in either of the equation, we get $q=56$
The relation between temperature scale A and B becomes
${{T}_{A}}=\dfrac{7}{5}{{T}_{B}}+56$
If temperature noted by hypothetical scale B is ${{60}^{{}^\circ }}$, then temperature observed by scale A is
${{T}_{A}}=\dfrac{7}{5}\times 60+56$
${{T}_{A}}={{140}^{{}^\circ }}$
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.
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