Chapter 1, Problem 86QP

If the temperature of a cup of coffee decreases from ${60.0}^{\circ }\text{C}\text{to}{25.0}^{\circ }\text{C}$ , what is the decrease in temperature in units of degrees Celsius and kelvin?

Interpretation Introduction

Interpretation:

The temperature change from $60°\text{C}$ to $\text{25 }°\text{C}$ in terms of Celsius and kelvin is to be determined.

Concept Introduction:

The temperature can be measured in terms of degree Celsius and kelvin. Temperature is an intensive property.

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

Answer to Problem 86QP

Solution:

The change in temperature from $60°\text{C}$ to $\text{25 }°\text{C}$ in terms of Celsius is $\text{35 }°\text{C}$ and in terms of kelvin is $\text{35 K}$ .

Explanation of Solution

:

The calculation for the change in temperature from $60°\text{C}$ to $\text{25 }°\text{C}$ is done by using the following relationship.

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

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

The initial temperature is given as $60°\text{C}$ and the final temperature is given as $\text{25 }°\text{C}$ .

Substitute $T_{°\text{C}1}$ as $60°\text{C}$ and $T_{°\text{C}2}$ as $\text{25 }°\text{C}$ in equation $\left(1\right)$ .

$\begin{array}{c}{T}_{°\text{C}}=60°\text{C}-25°\text{C}\\ =35°\text{C}\end{array}$

Therefore, the change in temperature $\left(T_{°\text{C}}\right)$ is calculated as $\text{35 }°\text{C}$ .

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 as $60°\text{C}$ . Substitute the value of $T_{°\text{C}1}$ in equation $\left(2\right)$ .

$\begin{array}{c}{T}_{\text{K1}}=60°\text{C}+273.15\\ =333.15\text{ K}\end{array}$

Here, the initial temperature $\left(T_{\text{K}1}\right)$ in the kelvin scale is calculated as $333.15\text{ K}$ .

The final temperature $\left(T_{°\text{C2}}\right)$ in $°\text{C}$ is given as $\text{25 }°\text{C}$ . Substitute the value of $T_{°\text{C}1}$ in equation $\left(2\right)$ .

$\begin{array}{c}{T}_{\text{K2}}=25°\text{C}+273.15\\ =298.15\text{ K}\end{array}$

Here, the final temperature $\left(T_{\text{K}2}\right)$ in the Kelvin scale is calculated as $298.15\text{ K}$ .

The calculation for the change in temperature from $298.15\text{ K}$ to $333.15\text{ K}$ is done by using the following relationship.

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

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

The initial temperature is $298.15\text{ K}$ and the final temperature is $333.15\text{ K}$ .

Substitute, $T_{\text{K}1}$ as $298.15\text{ K}$ and $T_{\text{K}2}$ as $333.15\text{ K}$ in equation $\left(1\right)$ .

$\begin{array}{c}{T}_{\text{K}}=333.15\text{ K}-298.15\text{ K}\\ =35\text{ K}\end{array}$

Therefore, the change in temperature $\left(T_{\text{K}}\right)$ is calculated as $\text{35 K}$ .

Conclusion

: The change in temperature from $60°\text{C}$ to $\text{25 }°\text{C}$ in terms of Celsius is $\text{35 }°\text{C}$ and in terms of kelvin is $\text{35 K}$ .