Subscribe $4.99/month

Un-lock Verified Step-by-Step Experts Answers.

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

All the data tables that you may search for.

Need Help? We got you covered.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 2

Q. 2.14

A new scale N of temperature is devised in such a way that the freezing point of ice is 100°N and boiling point is 400°N. What is the temperature reading on this new scale when the temperature is 150°C? At what temperature both the Celsius and new temperature scale reading would be the same?

Step-by-Step

Verified Solution

Let               t°N = ax + b

So that        100 = ax_i+ b

400 = ax_s + b

\begin{aligned}a &=\frac{300}{x_s-x_i} \\b &=100-\frac{300 x_i}{x_s-x_i} \\t^{\circ} N &=\frac{300 x}{x_s-x_i}+100-\frac{300 x_i}{x_s-x_i} \\&=\frac{300\left(x-x_i\right)}{x_s-x_i}+100\end{aligned}

Also  t^{\circ} C =100 \frac{100\left(x-x_i\right)}{x_s-x_i}

\therefore  t°N = 3 t°C + 100

When         t°C = 150, then

t°N = 3 × 150 + 100 = 550

For              t°N = t°C

t°C = 3t°C + 100

t°C = – 50