Question

If a thermometer reads freezing point of water as ${20}^{\circ }C$ and boiling point at ${150}^{\circ }C$ how much thermometer read when the actual temperature is ${60}^{\circ }C$.

• A. ${98}^{\circ }C$
• B. ${110}^{\circ }C$
• C. ${40}^{\circ }C$
• D. ${60}^{\circ }C$

Answer 1

The correct option is A ${98}^{\circ }C$

We let the unknown degree scale be called $x^{\circ }$

Now we know that the freezing point$={20}^{\circ }x$

The boiling point$={150}^{\circ }x$

This also means that ${20}^{\circ }x=0^{\circ }$ C($\because$ on a Celcius Scale, the freezing point is $0^{\circ }$)

And ${150}^{\circ }x={100}^{\circ }$C (On a Celcius Scale, boiling point is ${100}^{\circ }$)

Also ${(150-20)}^{\circ }x={100}^{\circ }$C ($\because$ on a Celcius Scale ${100}^{\circ }-0^{\circ }={100}^{\circ }$)

So, if ${100}^{\circ }$C$={130}^{\circ }x$

Then $1^{\circ }$C$={1.3}^{\circ }x$

To go from $C$ degree to $x$, we need to use:

$x=1.3C+20$

And $C=\frac{(x-20)}{1.3}$

So if the temperature given is ${60}^{\circ }$C, then we calculate $x$ as:

$C=\frac{(x-20)}{1.3}$

$\Rightarrow 60=\frac{(X-20)}{1.3}$

$\Rightarrow 60\times 1.3=x-20$

$\Rightarrow 78=x-20$

$\Rightarrow x={98}^{\circ }$

So ${60}^{\circ }$C will be equal to ${98}^{\circ }$