Question

A certain liquid has a melting point of -50$^\circ C$ and a boiling point of 150$^\circ C$. A thermometer is designed with this liquid and its melting and boiling points are designated at 0$^\circ L$ and 100$^\circ L$. The melting and boiling points of water on this scale are

Answer 1

The melting and boiling points of water on the L scale are $25^\circ L$ and $75^\circ L$ respectively.

Answer 2

Let the liquid’s Celsius temperature, $T_C$, relate to the L scale reading, $T_L$, by a linear relation:

$$T_C = aT_L + b.$$

Using the given points:

• When $T_L = 0$, $T_C = -50$:

  $-50 = a(0) + b \Rightarrow b = -50$.

• When $T_L = 100$, $T_C = 150$:

  $150 = 100a - 50 \Rightarrow 100a = 200 \Rightarrow a = 2$.

Thus, the conversion is:

$$T_C = 2T_L - 50 \quad \Rightarrow \quad T_L = \frac{T_C + 50}{2}.$$

For water:

• Melting point ($T_C = 0^\circ$C): $$T_L = \frac{0 + 50}{2} = 25^\circ L.$$
• Boiling point ($T_C = 100^\circ$C): $$T_L = \frac{100 + 50}{2} = 75^\circ L.$$