Suppose that on a linear temperature scale X, water boils at `-72.0^@X` and freezes at `-123.0^@X` . What is a temperature of 59.0 K on the X scale ? (Approximate water's boiling point as 373K .)

Answer

To find the temperature of 59.0 K on the X scale, we will follow these steps:

### Step 1: Identify the reference points
We know the following reference points:
- On the X scale:
- Water boils at \( -72.0^\circ X \)
- Water freezes at \( -123.0^\circ X \)

- On the Kelvin scale:
- Water boils at \( 373 \, K \)
- Water freezes at \( 273 \, K \)

### Step 2: Calculate the range of temperatures on both scales
- The range on the X scale:
\[ \text{Range}_X = -72 - (-123) = -72 + 123 = 51^\circ X \]

- The range on the Kelvin scale:
\[ \text{Range}_{K} = 373 - 273 = 100 \, K \]

### Step 3: Set up the ratio of temperature differences
Since the temperature intervals are proportional, we can set up the following ratio:
\[ \frac{T - (-123)}{51} = \frac{59 - 273}{100} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
\[ \frac{T + 123}{51} = \frac{59 - 273}{100} \]
Calculating \( 59 - 273 \):
\[ 59 - 273 = -214 \]
So the equation becomes:
\[ \frac{T + 123}{51} = \frac{-214}{100} \]

### Step 5: Cross-multiply to solve for T
Cross-multiplying gives us:
\[ 100(T + 123) = -214 \times 51 \]
Calculating \( -214 \times 51 \):
\[ -214 \times 51 = -10914 \]
Thus, we have:
\[ 100T + 12300 = -10914 \]

### Step 6: Isolate T
Now, isolate \( T \):
\[ 100T = -10914 - 12300 \]
Calculating the right side:
\[ -10914 - 12300 = -23214 \]
So:
\[ 100T = -23214 \]
Dividing by 100:
\[ T = -232.14^\circ X \]

### Conclusion
The temperature of \( 59.0 \, K \) on the X scale is approximately:
\[ T \approx -232.14^\circ X \]