Assume the thermometer scale is linear

Let the thermometer reading be $x$, and the actual temperature in degree Celsius be $T$.

Since the scale is linear, $x = aT + b$ where $a$ and $b$ are constants.

Use the given calibration points

• At ice point: $T=0^\circ C$, reading is $\dfrac{x_0}{3}$ $\frac{x_0}{3} = a(0) + b$ $b = \frac{x_0}{3}$

• At boiling point: $T=100^\circ C$, reading is $x_0$ $x_0 = 100a + \frac{x_0}{3}$

So, $100a = x_0 - \frac{x_0}{3} = \frac{2x_0}{3}$ $a = \frac{2x_0}{300} = \frac{x_0}{150}$

Now use the object reading

For the object, thermometer reading is $\dfrac{x_0}{2}$.

Hence, $\frac{x_0}{2} = aT + b$ $\frac{x_0}{2} = \frac{x_0}{150}T + \frac{x_0}{3}$

Solve for $T$

Subtract $\dfrac{x_0}{3}$ from both sides: $\frac{x_0}{2} - \frac{x_0}{3} = \frac{x_0}{150}T$

$x_0\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{x_0}{150}T$

$x_0\left(\frac{1}{6}\right)=\frac{x_0}{150}T$

Cancel $x_0$: $\frac{1}{6} = \frac{T}{150}$

Therefore, $T = 150 \cdot \frac{1}{6} = 25^\circ C$

Check options

The correct option is: $\boxed{\text{C: }25}$

Comparison with stored answer

Stored correct answer: C

Our derived answer is also C, so they agree.