Example 12 - The Fahrenheit F and absolute temperature K - Examples

  1. Chapter 10 Class 11 Straight Lines
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Example 12 (Method 1) The Fahrenheit temperature F and absolute temperature K satisfy a linear equation. Given that K = 273 when F = 32 and that K = 373 when F = 212. Express K in terms of F and find the value of F, when K = 0. Assuming F along x-axis & K along y-axis Given that K = 273 when F = 32 & K = 373 when F = 212 We have two points (32, 273) & (212, 373) in XY plane By two point form Equation on line passing through points (x1, y1) & (x2, y2) is (y – y1) = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) (x – x1) Here the point (F, K) is in the equation and line passes through points (273, 32) & (373, 212) So, our equation becomes (K – 273) = (373 − 273)/(212 − 32) (F – 32) K – 273 = 100/180 (F – 32) K – 273 = 5/9 (F – 32) K = 5/9 (F – 32) + 273 which is the required equation Now, finding the value of F, when K = 0. Putting K = 0 in the equation K = 5/9 (F – 32) + 273 0 = 5/9 (F – 32 ) + 273 0 – 273 = 5/9 (F – 32 ) -2-73 = 5/9 (F – 32 ) (−273 × 9)/5 = F – 32 (−273 × 9)/5 + 32 = F (−273 × 9 + 32 × 5 )/5 = F (−2457 + 160 )/5 = F (−2297)/5 = F F = (−2297)/5 F = – 459.4 Hence , when K = 0, F = –459.4 Example 12 (Method 2) The Fahrenheit temperature F and absolute temperature K satisfy a linear equation. Given that K = 273 when F = 32 and that K = 373 when F = 212. Express K in terms of F and find the value of F, when K = 0. We know that the equation of a line is y = mx + c Assuming F along x-axis & K along y-axis i.e. y = K & x = F So, our equation becomes K = mF + c We need to find value of m & c Given that K = 273 when F = 32 Putting K = 273 & F = 32 in equation (1) 273 = 32m + c Given that K = 373 when F = 212 Putting K = 373 & F = 212 in equation (1) 373 = 212m + c Now solving (A) & (B) 273 = 32m + c …(A) 373 = 212m + c …(B) Solving (A) 273 = 32m + c c = 273 – 32m Putting value of c in (B) 373 = 212m + c 373 = 212m + 273 − 32m 373 – 273 = 212m − 2m 100 = 180m 100/180 = m 5/9 = m m = 5/9 Putting value of m in (A) 273 = 32m + c c = 273 – 32m c = 273 − 32/1 × 5/9 c = (273 × 9 − 32 × 5)/9 c = (2457 − 160)/9 c = 2297/9 Putting value of m & c in (1) K = mF + c K = 5/9 F + 2297/9 which is the required equation Now, finding the value of F, when K = 0. Putting K = 0 in the equation K = 5/9 F + 2297/9 0 = 5/9 F + 2297/9 (−2297)/9 = 5/9 F 5/9 F = (−2297)/9 F = 9/5 × (−2297)/9 F = ( − 2297)/5 F = – 459.4 Hence , when K = 0, F = –459.4

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