The boiling point of water on the Kelvin temperature scale is 373.15 K and the freezing temperature of water is 273.15 K

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The boiling point of water on the Kelvin temperature scale is 373.15 K and the freezing temperature of water is 273.15 K. A temperature scale, called the Reaumur (R) scale, is defined by identifying the freezing point of water as 0.0°R and the boiling point of water as 80.0°R. What is 62.3°R on the Kelvin scale? Use the graphing method (y=mx+b).

The Correct Answer and Explanation is:

To convert 62.3°R to the Kelvin scale, we need to establish a linear relationship between the Reaumur (R) and Kelvin (K) scales. We can do this using the formula for a straight line y=mx+by = mx + by=mx+b, where:

  • yyy represents the temperature in Kelvin (K),
  • xxx represents the temperature in Reaumur (R),
  • mmm is the slope of the line, and
  • bbb is the y-intercept.

Step 1: Find the slope (m) and y-intercept (b)

We know the following data points:

  • The freezing point of water: 0.0°R corresponds to 273.15 K.
  • The boiling point of water: 80.0°R corresponds to 373.15 K.

Using these two points, we can calculate the slope of the line mmm:m=373.15 K−273.15 K80.0 R−0.0 R=100.080.0=1.25 K per °Rm = \frac{373.15 \, \text{K} – 273.15 \, \text{K}}{80.0 \, \text{R} – 0.0 \, \text{R}} = \frac{100.0}{80.0} = 1.25 \, \text{K per °R}m=80.0R−0.0R373.15K−273.15K​=80.0100.0​=1.25K per °R

Now, we can find the y-intercept bbb. Using the point (0, 273.15):y=mx+b  ⟹  273.15=1.25(0)+b  ⟹  b=273.15y = mx + b \implies 273.15 = 1.25(0) + b \implies b = 273.15y=mx+b⟹273.15=1.25(0)+b⟹b=273.15

Thus, the equation relating Reaumur (R) to Kelvin (K) is:K=1.25R+273.15K = 1.25R + 273.15K=1.25R+273.15

Step 2: Convert 62.3°R to Kelvin

Now that we have the equation, we can convert 62.3°R to Kelvin:K=1.25(62.3)+273.15K = 1.25(62.3) + 273.15K=1.25(62.3)+273.15K=77.875+273.15=351.025 KK = 77.875 + 273.15 = 351.025 \, \text{K}K=77.875+273.15=351.025K

Final Answer:

The temperature of 62.3°R on the Kelvin scale is 351.025 K.

This method relies on the understanding of linear relationships between temperature scales, allowing for easy conversion when given the freezing and boiling points on both scales. The linear equation K=1.25R+273.15K = 1.25R + 273.15K=1.25R+273.15 efficiently relates Reaumur and Kelvin temperatures across the entire range.

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