
A faulty thermometer reads ${5^ \circ }C$ in melting ice and ${99^ \circ }C$ in dry steam. Find the correct temperature in Fahrenheit scale if this faulty thermometer reads ${52^ \circ }C$.
A) ${84^ \circ }C$
B) ${122^ \circ }C$
C) ${56^ \circ }C$
D) ${88^ \circ }C$
Answer
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Hint: Use the fault reading of the melting ice, dry steam and the given temperature in the formula of the correct reading to find the correct reading in the Celsius scale. Convert the Celsius temperature into the Fahrenheit temperature using the formula.
Formula used:
(1) Formula of the correct reading is given by
$c = \dfrac{{f - m}}{{d - m}} \times 100$
Where $c$ is the correct reading, $f$ is the faulty reading, $d$ is the reading of the dry steam and $m$ is the reading of the melting ice in the faulty thermometer.
(2) The relation between the temperature in the Fahrenheit and the Celsius scale is given by
$\dfrac{C}{5} = \dfrac{{F - 32}}{9}$
Where $C$ is the temperature in the Celsius scale and $F$ is the temperature in the Fahrenheit scale.
Complete step by step solution:
It is given that the
Reading of the melting ice in the faulty thermometer, $m = {5^ \circ }C$
Reading of the dry steam in the faulty thermometer, $d = {99^ \circ }C$
Using the formula of the correct temperature,
$c = \dfrac{{f - m}}{{d - m}} \times 100$
Substituting the known values in the above formula, we get
$\Rightarrow c = \dfrac{{52 - 5}}{{99 - 5}} \times 100$
By performing various arithmetic operations, we get
$\Rightarrow c = {50^ \circ }C$
Using the relation between the Fahrenheit and the Celsius scale,
$\Rightarrow \dfrac{C}{5} = \dfrac{{F - 32}}{9}$
Substitute the temperature in the Celsius,
$\Rightarrow \dfrac{{50}}{5} = \dfrac{{F - 32}}{9}$
By simplification of the above equation, we get
$\Rightarrow F = 90 + 32 = {122^ \circ }F$
Hence the correct reading of the temperature in the Fahrenheit scale is obtained as ${122^ \circ }F$ .
Thus the option (B) is correct.
Note: At a normal condition, the temperature of the melting ice is ${0^ \circ }C$ , it can be said in the Fahrenheit as ${32^ \circ }F$ . Drying steam is obtained when the water is boiled and evaporated. Temperature of the drying steam is ${140^ \circ }F - {180^ \circ }F$ . Remember the formula for the correct reading and the Celsius and the Fahrenheit conversion.
Formula used:
(1) Formula of the correct reading is given by
$c = \dfrac{{f - m}}{{d - m}} \times 100$
Where $c$ is the correct reading, $f$ is the faulty reading, $d$ is the reading of the dry steam and $m$ is the reading of the melting ice in the faulty thermometer.
(2) The relation between the temperature in the Fahrenheit and the Celsius scale is given by
$\dfrac{C}{5} = \dfrac{{F - 32}}{9}$
Where $C$ is the temperature in the Celsius scale and $F$ is the temperature in the Fahrenheit scale.
Complete step by step solution:
It is given that the
Reading of the melting ice in the faulty thermometer, $m = {5^ \circ }C$
Reading of the dry steam in the faulty thermometer, $d = {99^ \circ }C$
Using the formula of the correct temperature,
$c = \dfrac{{f - m}}{{d - m}} \times 100$
Substituting the known values in the above formula, we get
$\Rightarrow c = \dfrac{{52 - 5}}{{99 - 5}} \times 100$
By performing various arithmetic operations, we get
$\Rightarrow c = {50^ \circ }C$
Using the relation between the Fahrenheit and the Celsius scale,
$\Rightarrow \dfrac{C}{5} = \dfrac{{F - 32}}{9}$
Substitute the temperature in the Celsius,
$\Rightarrow \dfrac{{50}}{5} = \dfrac{{F - 32}}{9}$
By simplification of the above equation, we get
$\Rightarrow F = 90 + 32 = {122^ \circ }F$
Hence the correct reading of the temperature in the Fahrenheit scale is obtained as ${122^ \circ }F$ .
Thus the option (B) is correct.
Note: At a normal condition, the temperature of the melting ice is ${0^ \circ }C$ , it can be said in the Fahrenheit as ${32^ \circ }F$ . Drying steam is obtained when the water is boiled and evaporated. Temperature of the drying steam is ${140^ \circ }F - {180^ \circ }F$ . Remember the formula for the correct reading and the Celsius and the Fahrenheit conversion.
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