Question 271: Thermodynamics & KTG

Question

On a temperature scale '$$\mathrm{X}$$', the boiling point of water is $$65^{\circ} \mathrm{X}$$ and the freezing point is $$-15^{\circ} \mathrm{X}$$. Assume that the $$\mathrm{X}$$ scale is linear. The equivalent temperature corresponding to $$-95^{\circ} \mathrm{X}$$ on the Farenheit scale would be:

Accepted Answer

We are given two temperature scales: the X-scale and the Celsius scale. The relationship between two linear temperature scales can be expressed as follows:
$$\frac{X - X_{	ext{freeze}}}{X_{	ext{boil}} - X_{	ext{freeze}}} = \frac{C - C_{	ext{freeze}}}{C_{	ext{boil}} - C_{	ext{freeze}}}$$
Here, $$X_{	ext{freeze}}$$ and $$X_{	ext{boil}}$$ are the freezing and boiling points of water on the X-scale, while $$C_{	ext{freeze}}$$ and $$C_{	ext{boil}}$$ are the freezing and boiling points of water on the Celsius scale.
We are given the following values:
X-scale:
$$X_{	ext{boil}} = 65^{\circ} \mathrm{X}$$
$$X_{	ext{freeze}} = -15^{\circ} \mathrm{X}$$
Celsius scale:
$$C_{	ext{boil}} = 100^{\circ} \mathrm{C}$$
$$C_{	ext{freeze}} = 0^{\circ} \mathrm{C}$$
Our goal is to find the equivalent temperature of $$-95^{\circ} \mathrm{X}$$ on the Fahrenheit scale.
Step 1: Convert $$-95^{\circ} \mathrm{X}$$ to Celsius:
Use the relationship between X-scale and Celsius scale:
$$\frac{-95 - (-15)}{65 - (-15)} = \frac{C - 0}{100 - 0}$$
Simplify and solve for C:
$$\frac{-80}{80} = \frac{C}{100}$$
$$C = -100^{\circ} \mathrm{C}$$
Step 2: Convert $$-100^{\circ} \mathrm{C}$$ to Fahrenheit:
Use the conversion formula between Celsius and Fahrenheit:
$$T_F = \frac{9}{5}T_C + 32$$
Substitute the Celsius temperature:
$$T_F = \frac{9}{5} 	imes (-100) + 32$$
$$T_F = -180 + 32$$
$$T_F = -148^{\circ} \mathrm{F}$$
So, the equivalent temperature corresponding to $$-95^{\circ} \mathrm{X}$$ on the Fahrenheit scale is $$-148^{\circ} \mathrm{F}$$.

Answer

$$-148^{\circ} \mathrm{F}$$

Answer

$$-48^{\circ} \mathrm{F}$$

Answer

$$-63^{\circ} \mathrm{F}$$

Answer

$$-112^{\circ} \mathrm{F}$$

Thermodynamics & KTG

On a temperature scale '$\mathrm{X}', the boiling point of water is 65^{\circ} \mathrm{X} and the freezing point is -15^{\circ} \mathrm{X}. Assume that the \mathrm{X} scale is linear. The equivalent temperature corresponding to -95^{\circ} \mathrm{X}$ on the Farenheit scale would be:

Choose the correct answer: