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Question
The following data gives the melting point of a alloy of lead and zinc where ‘t’ is the temperature in degree c and P is the percentage of lead in the alloy.
| P | 40 | 50 | 60 | 70 | 80 | 90 |
| T | 180 | 204 | 226 | 250 | 276 | 304 |
Find the melting point of the alloy containing 84 percent lead.
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Solution
Since the required value is at the end of the table
Apply backward interpolation formula.
To find T at p = 84
`"T"_(("p" = "p"_0 + "nh")) = "T"_"n" + "n"/(1!) ∇"T"_"n" + ("n"("n" + 1))/(2!) ∇^2"T"_0 + ("n"("n" + 1)("n" 2))/(3!) Delta^3"T"_0 + .......`
To find T at P = 84
Pn + nh = 84
90 + n(10) = 84
10n = 84 – 90
10n = – 6
⇒ n = `(-6)/10`
n = – 0.6
| P | T | `Delta"T"` | `Delta^2"T"` | `Delta^3"T"` | `Delta^4"T"` | `Delta^5"T"` |
| 40 | 180 | |||||
| 24 | ||||||
| 50 | 204 | – 2 | ||||
| 22 | 4 | |||||
| 60 | 226 | 2 | – 4 | |||
| 24 | 0 | 4 | ||||
| 70 | 250 | 2 | 0 | |||
| 26 | 0 | |||||
| 80 | 276 | 2 | ||||
| 28 | ||||||
| 90 | 304 |
`"T"_((84)) = 304 + ((-0.6))/(1!) (28) + ((-0.6)(-0.6 + 1))/(2!) (2)+ ((-0.6)(-0.6 + 1)(-0.6 + 2))/(3!) (0) + ((-0.6)(-0.6 + 1)(-0.6 + 2)(0.6 + 3))/(4!) (0) + ((-0.6)(-0.6 + 1)(-0.6 +2)(-0.6 + 3)(-0.6 + 4))/(5!) (4) +`
= 304 – 16.8 – 0.24 – 0.091392
= 304 – 17.131392
= 286.86
Hence the melting point of the alloy is 286.86°c.
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