Question

Find the value of 50 degree celsius on a scale where bp is 200 degree kelvin and freezing point is 20 degree kelvin

Answer 1

110

Answer 2

The relationship between two linear temperature scales is given by the formula: $\frac{T_1 - FP_1}{BP_1 - FP_1} = \frac{T_2 - FP_2}{BP_2 - FP_2}$ where $T$ is the temperature, $FP$ is the freezing point, and $BP$ is the boiling point of water on the respective scales (Scale 1 and Scale 2).

Let Scale 1 be the Celsius scale and Scale 2 be the new custom scale.

On the Celsius scale:

• Freezing point of water ($FP_C$) = $0^\circ C$
• Boiling point of water ($BP_C$) = $100^\circ C$
• Given temperature on Celsius scale ($T_C$) = $50^\circ C$

On the new custom scale:

• Freezing point of water ($FP_{new}$) = $20$ Kelvin
• Boiling point of water ($BP_{new}$) = $200$ Kelvin
• Let the equivalent temperature on the new scale be $T_{new}$.

Using the conversion formula: $\frac{T_C - FP_C}{BP_C - FP_C} = \frac{T_{new} - FP_{new}}{BP_{new} - FP_{new}}$ Substitute the known values: $\frac{50 - 0}{100 - 0} = \frac{T_{new} - 20}{200 - 20}$ $\frac{50}{100} = \frac{T_{new} - 20}{180}$ Simplify the left side: $\frac{1}{2} = \frac{T_{new} - 20}{180}$ Multiply both sides by 180: $\frac{1}{2} \times 180 = T_{new} - 20$ $90 = T_{new} - 20$ Add 20 to both sides to solve for $T_{new}$: $T_{new} = 90 + 20$ $T_{new} = 110$ The value of 50 degrees Celsius on the new scale is 110.