Chapter 19, Problem 1PQ

(a)

To determine

Convert temperature from Fahrenheit to Celsius and Kelvin.

Answer to Problem 1PQ

The temperature in Celsius is $\underset{\_}{24\text{°C}}$ and the temperature in Kelvin is $\underset{\_}{\text{297}\text{K}}$.

Explanation of Solution

Write the equation of temperature in centigrade scale.

  $T\left(\text{°C}\right)=\frac{5}{9}\left[T\left(\text{°F}\right)-32\right]$                                                                                    (I)

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

Write the expression for the temperature in Kelvin scale.

  $T\left(\text{K}\right)=T\left(\text{°C}\right)+273.15$                                                                                   (II)

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

Conclusion:

Substitute $76\text{°F}$ for $T\left(\text{°F}\right)$ in equation (I) to find $T\left(\text{°C}\right)$.

  $\begin{array}{c}T\left(\text{°C}\right)=\frac{5}{9}\left[\left(76\text{°F}\right)-32\right]\\ =\frac{5}{9}\left(44\right)\\ =24\text{°C}\end{array}$

Thus, The temperature in Celsius is $\underset{\_}{24\text{°C}}$.

Substitute $\text{24°C}$ for $T\left(\text{°C}\right)$ in equation (II) to find $T\left(\text{°K}\right)$.

  $\begin{array}{c}T\left(\text{K}\right)=\left(\text{24°C}\right)+273.15\\ =297\text{K}\end{array}$

Thus, the temperature in Kelvin is $\underset{\_}{\text{297}\text{K}}$.

(b)

To determine

Convert temperature from Fahrenheit to Celsius and Kelvin.

Answer to Problem 1PQ

For $180°\text{F}$.

The temperature in Celsius is $\underset{\_}{\text{82°C}}$ and the temperature in Kelvin is $\underset{\_}{355\text{K}}$.

For $190°\text{F}$.

The temperature in Celsius is $\underset{\_}{\text{88°C}}$ and the temperature in Kelvin is $\underset{\_}{361\text{K}}$.

Explanation of Solution

Write the equation of temperature in centigrade scale.

  $T\left(\text{°C}\right)=\frac{5}{9}\left[T\left(\text{°F}\right)-32\right]$                                                                                    (I)

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

Write the expression for the temperature in Kelvin scale.

  $T\left(\text{K}\right)=T\left(\text{°C}\right)+273.15$                                                                                   (II)

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

Conclusion:

Substitute $\text{180°F}$ for $T\left(\text{°F}\right)$ in equation (I) to find $T\left(\text{°C}\right)$.

  $\begin{array}{c}T\left(\text{°C}\right)=\frac{5}{9}\left[\left(\text{180°F}\right)-32\right]\\ =\frac{5}{9}\left(148\right)\\ =82\text{°C}\end{array}$

Substitute $\text{82°C}$ for $T\left(\text{°C}\right)$ in equation (II) to find $T\left(\text{°K}\right)$.

  $\begin{array}{c}T\left(\text{K}\right)=\left(\text{82°C}\right)+273.15\\ =355\text{K}\end{array}$

Thus, for $180°\text{F}$.

The temperature in Celsius is $\underset{\_}{\text{82°C}}$ and the temperature in Kelvin is $\underset{\_}{355\text{K}}$.

Substitute $\text{190°F}$ for $T\left(\text{°F}\right)$ in equation (I) to find $T\left(\text{°C}\right)$.

  $\begin{array}{c}T\left(\text{°C}\right)=\frac{5}{9}\left[\left(\text{190°F}\right)-32\right]\\ =\frac{5}{9}\left(158\right)\\ =88\text{°C}\end{array}$

Substitute $\text{88°C}$ for $T\left(\text{°C}\right)$ in equation (II) to find $T\left(\text{°K}\right)$.

  $\begin{array}{c}T\left(\text{K}\right)=\left(\text{88°C}\right)+273.15\\ =361\text{K}\end{array}$

Thus, for $190°\text{F}$.

The temperature in Celsius is $\underset{\_}{\text{88°C}}$ and the temperature in Kelvin is $\underset{\_}{361\text{K}}$.

(c)

To determine

Convert temperature from Fahrenheit to Celsius and Kelvin.

Answer to Problem 1PQ

The temperature in Celsius is $\underset{\_}{\text{-89°C}}$ and the temperature in Kelvin is $\underset{\_}{184\text{K}}$.

Explanation of Solution

Write the equation of temperature in centigrade scale.

  $T\left(\text{°C}\right)=\frac{5}{9}\left[T\left(\text{°F}\right)-32\right]$                                                                                    (I)

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

Write the expression for the temperature in Kelvin scale.

  $T\left(\text{K}\right)=T\left(\text{°C}\right)+273.15$                                                                                  (II)

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

Conclusion:

Substitute $\text{-129°F}$ for $T\left(\text{°F}\right)$ in equation (I) to find $T\left(\text{°C}\right)$.

  $\begin{array}{c}T\left(\text{°C}\right)=\frac{5}{9}\left[\left(\text{-129°F}\right)-32\right]\\ =\frac{5}{9}\left(-161\right)\\ =-89\text{°C}\end{array}$

Thus, the temperature in Celsius is $\underset{\_}{\text{-89°C}}$.

Substitute $-\text{89°C}$ for $T\left(\text{°C}\right)$ in equation (II) to find $T\left(\text{°K}\right)$.

  $\begin{array}{c}T\left(\text{K}\right)=\left(-\text{89°C}\right)+273.15\\ =184\text{K}\end{array}$

Thus, the temperature in Kelvin is $\underset{\_}{184\text{K}}$.