Chapter 19, Problem 5PQ

To determine

The values of the temperatures $T_1,T_2$ and $T_3$ in figure P19.4 on all three scales.

Answer to Problem 5PQ

The value of $T_1$ in Fahrenheit scale is $-40°\text{F}$ , in Celsius scale is $-40°\text{C}$ , in Kelvin scale is $233\text{ K}$ , value of $T_2$ in Fahrenheit scale is $0°\text{F}$ , in Celsius scale is $-18°\text{C}$ , in Kelvin scale is $255\text{ K}$ , value of $T_3$ in Fahrenheit scale is $32°\text{F}$ , in Celsius scale is $0°\text{C}$ , in Kelvin scale is $273.15\text{ K}$ ,.

Explanation of Solution

Write the relation between Celsius scale and Kelvin scale.

  $T\left(°\text{C}\right)=T\left(\text{K}\right)-273.15$                                                                                           (I)

Here, $T\left(°\text{C}\right)$ is the temperature in degree Celsius and $T\left(\text{K}\right)$ is the temperature in Kelvin scale.

Write the relation between Fahrenheit and Celsius scale.

  $T\left(°\text{F}\right)=\frac{9}{5}T\left(°\text{C}\right)+32$                                                                                             (II)

Here, $T\left(°\text{F}\right)$ is the temperature in Fahrenheit.

Write the general equation of straight-line in a $x-y$ graph.

  $y=mx+c$                                                                                                               (III)

Here, $y$ is the $y$ coordinate value, $x$ is the $x$ coordinate value, $m$ is the slope of straight-line and $c$ is the $y$ intercept.

The equation (II) gives the temperature in Fahrenheit in terms of temperature in degree Celsius.

In order to compare the slope of lines in the graph, substitute (I) in (II).

  $\begin{array}{c}T\left(°\text{F}\right)=\frac{9}{5}\left(T\left(\text{K}\right)-273.15\right)+32\\ =\frac{9}{5}T\left(\text{K}\right)-\frac{9}{5}\left(273.15\right)+32\\ =\frac{9}{5}T\left(\text{K}\right)-460\end{array}$                                                                        (IV)

Thus, comparing equations (I) and (III), it is evident that slope of Celsius versus Kelvin scale graph is 1 and $y$ intercept is $-273.15$ .

Comparing equations (IV) and (III), it is evident that slope of Fahrenheit versus Kelvin scale graph is $\frac{9}{5}$ and $y$ intercept is $-460$ .

Therefore, slope and $y$ intercept of straight-line in Fahrenheit versus Kelvin scale graph is greater than that in Celsius versus Kelvin scale graph.

In figureP19.4, slope and $y$ intercept of line A is greater than slope and $y$ intercept of line B.

Above explanations indicate that line A is Fahrenheit and B is Celsius.

Conclusion:

The temperature $T_1$ occurs when the lines A and B intersect or when temperatures on the Fahrenheit and Celsius scales are equal.

Substitute $T_1$ for $T\left(°\text{F}\right)$ and $T\left(°\text{C}\right)$ in equation (II) to find the value of $T_1$ .

  $\begin{array}{c}{T}_1=\frac{9}{5}{T}_1+32\\ T_1\left(1-\frac{9}{5}\right)=32\\ T_1=\frac{32}{\raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\\ =-40\end{array}$

$T_1$ is the same temperature in Fahrenheit and Celsius scales.

  $T_1=-40°\text{F or }-40°\text{C}$

Substitute $-40°\text{C}$ for $T\left(°\text{C}\right)$ in equation (I) to find the value of $T_1$ in Kelvin scale.

  $\begin{array}{c}-40°\text{C}=T_1\left(\text{K}\right)-273.15\\ T_1\left(\text{K}\right)=-40°\text{C}+273.15\\ =233.15\text{ K}\\ \approx \text{233 K}\end{array}$

$T_2$ is the temperature at which the line A or the Fahrenheit scale crosses zero.

  $T_2\left(°\text{F}\right)=0°\text{F}$

Substitute $0°\text{F}$ for $T\left(°\text{F}\right)$ in equation (II) to find the value of $T_2$ in Celsius scale.

  $\begin{array}{c}0°\text{F}=\frac{9}{5}{T}_2\left(°\text{C}\right)+32\\ T_2\left(°\text{C}\right)=\frac{-32}{9/5}\\ =-18°\text{C}\end{array}$

Substitute $-18°\text{C}$ for $T\left(°\text{C}\right)$ in equation (I) to find the value of $T_2$ in Kelvin scale.

  $\begin{array}{c}-18°\text{C}=T_2\left(\text{K}\right)-273.15\\ T_2\left(\text{K}\right)=-18°\text{C}+273.15\\ =255.15\text{ K}\\ \approx \text{255 K}\end{array}$

$T_3$ is the temperature at which the line B or the Celsius scale crosses zero.

  $T_3\left(°\text{C}\right)=0°\text{C}$

Substitute $0°\text{C}$ for $T\left(°\text{C}\right)$ in equation (II) to find the value of $T_3$ in Fahrenheit scale.

  $\begin{array}{c}{T}_3\left(°\text{F}\right)=\frac{9}{5}\left(0°\text{C}\right)+32\\ T_3\left(°\text{F}\right)=32°\text{F}\end{array}$

Substitute $0°\text{C}$ for $T\left(°\text{C}\right)$ in equation (I) to find the value of $T_3$ in Kelvin scale.

  $\begin{array}{c}0°\text{C}=T_3\left(\text{K}\right)-273.15\\ T_3\left(\text{K}\right)=0°\text{C}+273.15\\ =273.15\text{ K}\end{array}$

Therefore, the value of $T_1$ in Fahrenheit scale is $-40°\text{F}$ , in Celsius scale is $-40°\text{C}$ , in Kelvin scale is $233\text{ K}$ , value of $T_2$ in Fahrenheit scale is $0°\text{F}$ , in Celsius scale is $-18°\text{C}$ , in Kelvin scale is $255\text{ K}$ , value of $T_3$ in Fahrenheit scale is $32°\text{F}$ , in Celsius scale is $0°\text{C}$ , in Kelvin scale is $273.15\text{ K}$ ,.