Question

The body temperature of a patient is ${5.4^0}$ F above the normal temperature. His body temperature now in ${0^{0}}$C is
A: ${38^0}$C
B: ${98^{0}}$C
C: ${50^{0}}$ C
D: ${40^{0}}$C

Answer 1

Hint: We are given the body temperature of a patient in Fahrenheit which is above the normal temperature we have to find the body temperature now of his body in ${0^0}C$ the average normal body temperature is generally accepted as ${98.6^0}F$ studies also shows that the normal body temperature can have a wide range from ${97^0}F$$\left( {{{36.1}^0}c} \right)$ to ${99^0}F$$\left[ {{{37.2}^{0}}C} \right]$ the temperature given to us in Fahrenheit is the temperature which is above the body temperature so firstly we will add the body temperature into it so that we can find the actual temperature then we will convert into the degree Celsius

Formula used: $C = \dfrac{5}{9}\left( {F - {{32}^{0}}} \right)$
Here F= temperature in Fahrenheit

Complete step-by-step answer:
STEP1: we are given a temperature of ${5.4^0}F$ which is above the normal body temperature. Firstly we will find the exact temperature. In fahrenheit the normal body temperature is considered as ${98.6^0}F$ on adding this temperature to body temperature we will get the exact temperature so
F$ = $ Temperature given $ + {98.6^0}$F
F=$5.4 + {98.6^{0}}$
On further adding
F=$104$ degree Fahrenheit
STEP2: Now using the formula of conversion to degree Celsius we can use this formula
C=$\dfrac{5}{9}\left( {F - 32} \right)$
On substituting the value of F in this we get
STEP3: C$ = \dfrac{5}{9}\left( {104 - 32} \right)$
On Subtracting we get:
C$ = \dfrac{5}{9} \times 72$
On further solving:
C$ = 5 \times 8$
C$ = 40$

Temperature in ${0^0}C$ is ${40^0}$ hence option $\left( D \right)$ is the correct answer

Note: In this type of question students mainly confused the temperature above the body temperature as exact temperature and solve the whole question by using that temperature they should understand that the temperature is above normal body temperature so we have to add ${98.6^0}F$ i.e. normal body temperature in it. Some studies have shown that the “normal” body temperature can have a wide range, from ${97^0}$ F $({36.1^0}C)$ to ${99^0}$.
The fahrenheit to Celsius conversion equation may be expressed in this form:
${F^0} = 1.8{(^0}C) + 32$
Enter $98.6$ for the F0
$98.6 = 1.8{(^0}C) + 32$
$1.8{(^0}C) = 98.6 - 32$
$1.8{(^0}C) = 66.6$
0C$ = \dfrac{{66.6}}{{1.8}}$
0C$ = 37.0$
To solve for kelvin:
K=0C$ + 273$
K$ = 37.0 + 273$
K=$310$