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The boiling point of water decreases with altitude. At an altitude ℎ metres above sealevel, the boiling point of water is approximately 100 — 𝑇 degrees C where 𝑇 satisfiesthe equation 𝑇2 + 244𝑇 − ℎ = 0What is the boiling point when ℎ = 1500 metres?

Question

The boiling point of water decreases with altitude. At an altitude ℎ metres above sealevel, the boiling point of water is approximately 100 — 𝑇 degrees C where 𝑇 satisfiesthe equation 𝑇2 + 244𝑇 − ℎ = 0What is the boiling point when ℎ = 1500 metres?

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Solution

Para resolver el problema, primero necesitamos encontrar el valor de T T que satisface la ecuación cuadrática dada:

T2+244Th=0 T^2 + 244T - h = 0

Dado que h=1500 h = 1500 metros, sustituimos este valor en la ecuación:

T2+244T1500=0 T^2 + 244T - 1500 = 0

Ahora, resolvemos esta ecuación cuadrática usando la fórmula general para ecuaciones cuadráticas ax2+bx+c=0 ax^2 + bx + c = 0 :

T=b±b24ac2a T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

En este caso, a=1 a = 1 , b=244 b = 244 y c=1500 c = -1500 . Sustituimos estos valores en la fórmula:

T=244±244241(1500)21 T = \frac{-244 \pm \sqrt{244^2 - 4 \cdot 1 \cdot (-1500)}}{2 \cdot 1}

Calculamos el discriminante:

244241(1500)=59536+6000=65536 244^2 - 4 \cdot 1 \cdot (-1500) = 59536 + 6000 = 65536

Entonces, la ecuación se convierte en:

T=244±655362 T = \frac{-244 \pm \sqrt{65536}}{2}

Sabemos que 65536=256 \sqrt{65536} = 256 , así que:

T=244±2562 T = \frac{-244 \pm 256}{2}

Esto nos da dos soluciones:

T1=244+2562=122=6 T_1 = \frac{-244 + 256}{2} = \frac{12}{2} = 6

T2=2442562=5002=250 T_2 = \frac{-244 - 256}{2} = \frac{-500}{2} = -250

Dado que T T representa una temperatura, descartamos la solución negativa. Por lo tanto, T=6 T = 6 .

Finalmente, la temperatura de ebullición del agua a una altitud de 1500 metros es:

100T=1006=94 grados C 100 - T = 100 - 6 = 94 \text{ grados C}

Por lo tanto, la temperatura de ebullición del agua a 1500 metros de altitud es aproximadamente 94 grados Celsius.

This problem has been solved

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