Question

Heat stroke .If the body’s temperature is above $\mathbf{105}\mathbf{°}\mathbf{F}$ for a prolonged period, heat stroke can result. Express this temperature on the Celsius and Kelvin scales.

Short answer

The temperature in Celsius scale is $40.6°\mathrm{C}$

The temperature in Kelvin scale is $313.8\mathrm{K}$

Given :

The temperature in Fahrenheit scale is $105°\mathrm{F}$

Step-by-step solution

Step 1:

To covert the temperature on Fahrenheit scale into Celsius scale.

Temperature in Fahrenheit scale can be converted in temperature in Celsius scale by formula :

${\mathrm{T}}_{\mathrm{c}}=\frac{5}{9}\left({\mathrm{T}}_{\mathrm{F}}-32\right)$

Where is the ${\mathrm{T}}_{\mathrm{F}}$temperature in Fahrenheit scale and ${\mathrm{T}}_{\mathrm{C}}$temperature in Celsius scale.

To covert the temperature on Celsius scale to Kelvin scale

Temperature in Celsius scale can be converted in temperature in Kelvin scale by formula :

${\mathrm{T}}_{\mathrm{K}}={\mathrm{T}}_{\mathrm{c}}+273.15$

Where is the ${\mathrm{T}}_{\mathrm{c}}$temperature in Celsius scale ${\mathrm{T}}_{\mathrm{K}}$temperature is Kelvin scale.

Calculation of temperature in Celsius scale and Kelvin scale

For conversion of temperature from Fahrenheit scale to Celsius scale

Using formula

${\mathrm{T}}_{\mathrm{c}}=\frac{5}{9}\left({\mathrm{T}}_{\mathrm{F}}-32\right)$

Now putting the values of constants in above equation

$$\begin{gathered} {\mathrm{T}}_{\mathrm{c}}=\frac{5}{9}\left(105-32\right) \\ {\mathrm{T}}_{\mathrm{c}}=\frac{5}{9}\times \left(73\right) \\ {\mathrm{T}}_{\mathrm{c}}=40.6°\mathrm{C} \end{gathered}$$

Thus, the temperature in Celsius scale is $40.6°\mathrm{C}$.

For conversion of temperature from Celsius scale to Kelvin scale

Using formula

${\mathrm{T}}_{\mathrm{K}}={\mathrm{T}}_{\mathrm{c}}+273.15$

Now, putting the values of constant in above equation

$$\begin{gathered} {\mathrm{T}}_{\mathrm{K}}=40.6+273.15 \\ {\mathrm{T}}_{\mathrm{K}}=313.8\mathrm{K} \end{gathered}$$

This,the temperature in Kelvin scale is $313.8\mathrm{K}$