Question
The variation of the specific heat capacity of air with temperature is given in the following set of data:
| Heat Capacity (in kJ kg−1 K−1) |
1.003 |
1.005 |
1.008 |
1.013 |
1.020 |
1.029 |
| Temperature (in K) | 250 | 300 | 350 | 400 | 450 | 500 |
Compute the correlation coefficient rXY.
Answer :
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Skip in 10
We are asked to compute the correlation coefficient $r_{XY}$ between temperature $T$ (X) and specific heat capacity $C_p$ (Y) using the given data. Let's solve it manually, step by step. --- ### Step 1: Tabulate the data | $X = T$ (K) | 250 | 300 | 350 | 400 | 450 | 500 | | ------------------- | ----- | ----- | ----- | ----- | ----- | ----- | | $Y = C_p$ (kJ/kg·K) | 1.003 | 1.005 | 1.008 | 1.013 | 1.020 | 1.029 | --- ### Step 2: Compute sums needed for correlation coefficient The formula for correlation coefficient $r$ is: $$ r = \frac{n\sum XY - (\sum X)(\sum Y)}{\sqrt{\left[n\sum X^2 - (\sum X)^2\right]\left[n\sum Y^2 - (\sum Y)^2\right]}} $$ where $n = 6$ (number of data points). --- ### Step 3: Compute $\sum X, \sum Y, \sum X^2, \sum Y^2, \sum XY$ Compute $\sum X$ and $\sum Y$: $$ \sum X = 250 + 300 + 350 + 400 + 450 + 500 = 2250 $$ $$ \sum Y = 1.003 + 1.005 + 1.008 + 1.013 + 1.020 + 1.029 $$ Step by step: 1.003 + 1.005 = 2.008 2.008 + 1.008 = 3.016 3.016 + 1.013 = 4.029 4.029 + 1.020 = 5.049 5.049 + 1.029 = 6.078 $\sum Y = 6.078$ --- Compute $\sum
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