Question

Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 15.1 K.

Part A:
What is the magnitude of its temperature change (Change in T = 15.1 K) in degrees Celsius?
Answer: 15.1 C

Part B:
What is the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit?

Answer 1

Concepts and reason

The concepts used to solve this problem is temperature conversion between Kelvin to Celsius and Kelvin to Fahrenheit.

Use the conservation of units of temperature to convert the change in temperature from Kelvin to Celsius and Kelvin to Fahrenheit.

Fundamentals

Temperature defines the intensity or the degree of heat present in an object or substance.

The expression for the conversion of temperature from Kelvin to Celsius is,

${T_{\rm{C}}} = {T_{\rm{K}}} - 273$

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

The expression for the conversion of temperature from Celsius to Fahrenheit is,

${T_{\rm{F}}} = 1.8{T_{\rm{C}}} + 32$

Here, ${T_{\rm{F}}}$ is the temperature in Fahrenheit.

(Part A)

The expression for the conversion of temperature from Kelvin to Celsius is,

${T_{\rm{C}}} = {T_{\rm{K}}} - 273$

The expression for the change in temperature is,

$\left| {\Delta {T_{\rm{C}}}} \right| = \left| {{T_{{\rm{Ci}}}} - {T_{{\rm{Cf}}}}} \right|$

Here, ${T_{{\rm{Ci}}}}$ is the initial temperature in Celsius and ${T_{{\rm{Cf}}}}$ is the final temperature in Celsius.

Substitute ${T_{{\rm{Ki}}}} - 273$ for ${T_{{\rm{Ci}}}}$ and ${T_{{\rm{Kf}}}} - 273$ for ${T_{{\rm{Cf}}}}$ to find $\left| {\Delta {T_{\rm{C}}}} \right|$ .

$\begin{array}{c}\\\left| {\Delta {T_{\rm{C}}}} \right| = \left| {\left( {{T_{{\rm{Ki}}}} - 273} \right) - \left( {{T_{{\rm{Kf}}}} - 273} \right)} \right|\\\\ = \left| {{T_{{\rm{Ki}}}} - {T_{{\rm{Kf}}}}} \right|\\\\ = \left| {\Delta {T_{\rm{K}}}} \right|\\\end{array}$

Here, $\left| {\Delta {T_{\rm{K}}}} \right|$ is the change in temperature in kelvin.

The change in temperature is 15.1 K. In Kelvin scale and Celsius scale for the freezing point and boiling point of water there is 100 degree-intervals in both the scale.

Thus, $\left| {\Delta {T_{\rm{C}}}} \right| = \left| {\Delta {T_{\rm{K}}}} \right|$ .

Substitute $15.1$ for $\left| {{T_{{\rm{Ki}}}} - {T_{{\rm{Kf}}}}} \right|$ to find $\left| {\Delta {T_{\rm{C}}}} \right|$ .

$\Delta T = 15.1\,^\circ {\rm{C}}$

(Part B)

The expression for the conversion of temperature from Celsius to Fahrenheit is,

${T_{\rm{F}}} = 1.8{T_{\rm{C}}} + 32$

The expression for the magnitude of change in temperature in Fahrenheit is,

$\left| {\Delta {T_F}} \right| = \left| {{T_{{\rm{Fi}}}} - {T_{{\rm{Ff}}}}} \right|$

Here, ${T_{{\rm{Fi}}}}$ is the initial temperature in Fahrenheit and ${T_{{\rm{Ff}}}}$ is the final temperature in Fahrenheit.

Substitute $1.8{T_{{\rm{Ci}}}} + 32$ for ${T_{{\rm{Fi}}}}$ and $1.8{T_{{\rm{Cf}}}} + 32$ for ${T_{{\rm{Ff}}}}$ to find $\left| {\Delta {T_F}} \right|$ .

$\begin{array}{c}\\\left| {\Delta {T_{\rm{F}}}} \right| = \left| {\left( {1.8{T_{{\rm{Ci}}}} + 32} \right) - \left( {1.8{T_{{\rm{Cf}}}} + 32} \right)} \right|\\\\ = 1.8\left| {{T_{{\rm{Ci}}}} - {T_{{\rm{Cf}}}}} \right|\\\\ = 1.8\left| {\Delta {T_C}} \right|\\\end{array}$

The change in temperature in Kelvin scale is equal to the change in temperature in Celsius scale.

$\left| {\Delta {T_{\rm{C}}}} \right| = \left| {\Delta {T_{\rm{K}}}} \right|$

So, change of 15.1 K is equal to the change of $15.1\,^\circ {\rm{C}}$ .

Substitute 15.1 for $\left| {\Delta {T_C}} \right|$ in the above expression to find $\left| {\Delta {T_{\rm{F}}}} \right|$

$\begin{array}{c}\\\Delta {T_{\rm{F}}} = \left( {1.8} \right)\left( {15.1} \right)\\\\ = 27.18\,^\circ {\rm{F}}\\\\ \approx 27.2\,^\circ {\rm{F}}\\\end{array}$

Ans: Part A

The magnitude of its temperature change (Change in $T = 15.1\,{\rm{K}}$ ) in degree Celsius is $15.1\,^\circ {\rm{C}}$ .

Part B

The magnitude of its temperature change (Change in $T = 15.1\,{\rm{K}}$ ) in degree Fahrenheit is $27.2\,^\circ {\rm{F}}$ .