Question

The freezing point on a thermometer is marked as $20^\circ C$ and the boiling point as $150^\circ C$ a temperature of $60^\circ C$ on this thermometer will be read as
A. $40^\circ C$
B. $65^\circ C$
C. $98^\circ C$
D. $110^\circ C$

Answer 1

So in order to solve this question we need to understand what temperature is. Temperature is a device which measures temperature difference or temperature gradient. Boiling point of water and melting point or ice point of water is not fixed and it depends on altitude, latent heat. As due to altitude variation pressure varies so according to the Cauchy equation, boiling point of water rises at high altitude so it is difficult to cook at high altitudes. Also melting point of water, quartz crystal decreases with increase in pressure but for other substances melting point increases with increase in pressure.

Complete step by step answer:
Melting point on thermometer, ${T_1} = 20^\circ C$.
And boiling point, ${T_2} = 150^\circ C$.
So calibration on this scale is, $T = 150 - 20$.
$T = 130^\circ C$
But on a normal thermometer calibration is, $100^\circ C$. Let $t$ be temperature on this scale. So $60^\circ C$ on this scale can be found as,
$\dfrac{{t - 20}}{{130}} = \dfrac{{60 - 0}}{{100}} \to (i)$
Here, Melting point on this thermometer, ${T_1} = 20^\circ C$.
And Melting point on standard thermometer, ${T_O} = 0^\circ C$.
So, from equation (i) we get, $t - 20 = 78$.
$\therefore t = 98^\circ C$

So the correct option is C.

Note: It should be remembered that temperature can be measured in three scales, one is Celsius scale, other is Fahrenheit scale and the Kelvin scale. Kelvin is considered an absolute scale for measuring temperature and it is also the Standard International Unit of temperature measurement. Calibration of the thermometer is done mainly in cooking by two methods, one is freezing point method and other is by melting point method.