Chapter 1, Problem 124QP

If the temperature in a room increases from $\text{20}\text{.0°C to 30}\text{.0°C,}$ what is file temperature change in units of kelvin?

Interpretation Introduction

Interpretation:

The change in temperature in terms of Kelvin is to be calculated.

Concept Introduction:

Kelvin is the SI unit of temperature and is termed as the absolute temperature scale which depicts the lowest possible temperature in any system that is $0\text{ K}$ .

Celsius is the most common measuring unit for temperature. Celsius scale was defined by using freezing and boiling points of water but now, the triple point of water is introduced and the Boltzmann constant is the new basis for its definition.

Answer to Problem 124QP

Solution: The change in temperature in terms of kelvin is $10\text{ K}$ .

Explanation of Solution

The calculation for the change in temperature from $20°\text{C}$ to $30°\text{C}$ by using the following relationship is

$T_{°\text{C}}=T_{°\text{C}2}-T_{°\text{C}1}$ …… $\left(1\right)$

Here, $T_{°\text{C}}$ is the change in temperature in Celsius, $T_{°\text{C}2}$ is the final temperature, $T_{°\text{C}1}$ is the initial temperature.

The initial temperature is given as $20°\text{C}$ and the final temperature is given as $30°\text{C}$ .

Substitute $T_{°\text{C}1}$ as $20°\text{C}$ and $T_{°\text{C}2}$ as $30°\text{C}$ in equation $\left(1\right)$ .

$\begin{array}{c}{T}_{°\text{C}}=30°\text{C}-20°\text{C}\\ =10°\text{C}\end{array}$

Therefore, the change in temperature $\left(T_{°\text{C}}\right)$ is calculated as $10°\text{C}$ .

Now, the change in temperature in the Kelvin scale can be calculated as follows:

The conversion of Kelvin into degree Celsius can be carried out using the following relation:

$T_{\text{K}}=T_{°\text{C}}+273.15$

Here, $T_{\text{K}}$ is the temperature in Kelvin and $T_{°\text{C}}$ is the temperature in degree Celsius.

The initial temperature $\left(T_{°\text{C}1}\right)$ in $°\text{C}$ is given $20°\text{C}$ . Substitute the value of $T_{°\text{C}1}$ in equation $\left(2\right)$ .

$\begin{array}{c}{T}_{\text{K1}}=20°\text{C}+273.15\\ =293.15\text{ K}\end{array}$

Here, the initial temperature $\left(T_{\text{K}1}\right)$ in the Kelvin scale is calculated as $293.15\text{ K}$ .

The final temperature $\left(T_{°\text{C2}}\right)$ in $°\text{C}$ is given $30°\text{C}$ . Substitute the value of $T_{°\text{C}1}$ in equation $\left(2\right)$ .

$\begin{array}{c}{T}_{\text{K2}}=30°\text{C}+273.15\\ =303.15\text{ K}\end{array}$

Here, the final temperature $\left(T_{\text{K}2}\right)$ in the Kelvin scale is calculated as $303.15\text{ K}$ .

The calculation for the change in temperature from $293.15\text{ K}$ to $303.15\text{ K}$ by using the following relationship is

$T_{\text{K}}=T_{\text{K}2}-T_{\text{K}1}$ …… $\left(1\right)$

Here, $T_{\text{K}}$ is the change in temperature in Kelvin, $T_{\text{K}2}$ is the final temperature, $T_{\text{K}1}$ is the initial temperature.

The initial temperature is $293.15\text{ K}$ and the final temperature is $303.15\text{ K}$ .

Substitute $T_{\text{K}1}$ as $293.15\text{ K}$ and $T_{\text{K}2}$ as $303.15\text{ K}$ in equation $\left(1\right)$ .

$\begin{array}{c}{T}_{\text{K}}=303.15\text{ K}-293.15\text{ K}\\ =10\text{ K}\end{array}$

Therefore, the change in temperature $\left(T_{\text{K}}\right)$ is calculated as $10\text{ K}$ .

Therefore, the change in temperature in units of Kelvin is the same as the temperature change in units of Celsius.

Conclusion

The temperature change in units of Kelvin is the same as the temperature change obtained in units of Celsius.