On a linear temperature scale Y, water freezes at $$-160°$$Y and boils at $$-50°$$Y. On this Y scale, a temperature of 340 K would be read as : (water freezes at 273 K and boils at 373 K)

We are given two temperature scales: Kelvin (K) and a linear scale Y. The freezing and boiling points of water are known on both scales:

• On Kelvin scale: freezing point = 273 K, boiling point = 373 K
• On Y scale: freezing point = -160°Y, boiling point = -50°Y

We need to find the temperature on the Y scale that corresponds to 340 K. Since both scales are linear, the relationship between them can be expressed as:

$$T_Y = m \cdot T_K + c$$

where $$T_Y$$ is the temperature in °Y, $$T_K$$ is the temperature in K, $$m$$ is the slope, and $$c$$ is the intercept.

To find $$m$$ and $$c$$, we use the two known points:

• Point 1: (273 K, -160°Y)
• Point 2: (373 K, -50°Y)

The slope $$m$$ is calculated as the change in Y divided by the change in K:

$$m = \frac{\Delta T_Y}{\Delta T_K} = \frac{(-50) - (-160)}{373 - 273} = \frac{-50 + 160}{100} = \frac{110}{100} = 1.1$$

So, $$m = 1.1$$.

Now, substitute one point into the equation to find $$c$$. Using point 1 (273 K, -160°Y):

$$-160 = 1.1 \cdot 273 + c$$

First, compute $$1.1 \cdot 273$$:

$$1.1 \cdot 273 = 1.1 \cdot (200 + 73) = 1.1 \cdot 200 + 1.1 \cdot 73 = 220 + 80.3 = 300.3$$

So,

$$-160 = 300.3 + c$$

Solving for $$c$$:

$$c = -160 - 300.3 = -460.3$$

Thus, the equation is:

$$T_Y = 1.1 \cdot T_K - 460.3$$

Now, find $$T_Y$$ when $$T_K = 340$$ K:

$$T_Y = 1.1 \cdot 340 - 460.3$$

Compute $$1.1 \cdot 340$$:

$$1.1 \cdot 340 = 1.1 \cdot (300 + 40) = 1.1 \cdot 300 + 1.1 \cdot 40 = 330 + 44 = 374$$

Then,

$$T_Y = 374 - 460.3 = -86.3$$

Therefore, 340 K corresponds to -86.3°Y on the Y scale.

Comparing with the options:

• A: -73.7°Y
• B: -233.7°Y
• C: -86.3°Y
• D: -106.3°Y

Hence, the correct answer is Option C.