If a temperature scale defined the freezing point of water as `100^(@)` and the boiling point is `0^(@)`. What temperature on thils scale corresponds to `25^(@)C`?

Answer

To solve the problem of converting 25°C to the new temperature scale where the freezing point of water is defined as 100° and the boiling point as 0°, we can follow these steps:

### Step 1: Understand the new temperature scale
In the new scale:
- Freezing point of water (0°C) corresponds to 100° on the new scale.
- Boiling point of water (100°C) corresponds to 0° on the new scale.

### Step 2: Set up the conversion formula
We can use a linear transformation to convert between the Celsius scale and the new scale. The formula for conversion can be expressed as:
\[ K = \frac{(T - T_f)}{(T_b - T_f)} \times (K_b - K_f) + K_f \]
Where:
- \(K\) is the temperature in the new scale.
- \(T\) is the temperature in Celsius (25°C in this case).
- \(T_f\) is the freezing point in Celsius (0°C).
- \(T_b\) is the boiling point in Celsius (100°C).
- \(K_f\) is the freezing point in the new scale (100°).
- \(K_b\) is the boiling point in the new scale (0°).

### Step 3: Substitute the values
Substituting the values into the formula:
- \(T = 25°C\)
- \(T_f = 0°C\)
- \(T_b = 100°C\)
- \(K_f = 100°\)
- \(K_b = 0°\)

The equation becomes:
\[ K = \frac{(25 - 0)}{(100 - 0)} \times (0 - 100) + 100 \]

### Step 4: Simplify the equation
Now, simplify the equation:
\[ K = \frac{25}{100} \times (-100) + 100 \]
\[ K = 0.25 \times (-100) + 100 \]
\[ K = -25 + 100 \]
\[ K = 75° \]

### Final Answer
Thus, the temperature of 25°C corresponds to 75° on the new temperature scale.

---