contestada

Newtor's of cooling T = A * e ^ (- H) + C where is the temperature of the object at time and C is the constant temperature of the surrounding medium Supposem that the room temperature is 67 degrees and the temperature of a cup of coffee is 148 when it is placed on the table How long will it take for the coffee to cool to 102 for k=0.073354) Round your answer to two decimal places

Newtors of cooling T A e H C where is the temperature of the object at time and C is the constant temperature of the surrounding medium Supposem that the room t class=

Respuesta :

We are given that:

[tex]\begin{gathered} C=67^{\circ}, \\ k=0.073354. \end{gathered}[/tex]

We know that, at t=0,

[tex]T=148^{\circ}=Ae^0+C,[/tex]

solving the above equation for A, we get:

[tex]A=148^{\circ}-67^{\circ}=81^{\circ}.[/tex]

Now, setting T=102°, we get:

[tex]102^{\circ}=81^{\circ}e^{-kt}+67^{\circ}.[/tex]

Solving the above equation for t, we get:

[tex]\begin{gathered} 102^{\circ}-67^{\circ}=81^{\circ}e^{-0.073354t}, \\ 35^{\circ}=81^{\circ}e^{-0.073354t}, \\ e^{-0.073354t}=\frac{35}{81}, \\ -0.073354t=ln(\frac{35}{81}), \\ t=\frac{ln(\frac{35}{81})}{-0.073354}. \end{gathered}[/tex]

Finally, we get:

[tex]t\approx11.44.[/tex]

Answer:

[tex]11.44[/tex]

Otras preguntas