Question

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

Answer 1

We are given that:

`\begin{gathered} C=67^(\circ), \\ k=0.073354. \end{gathered}`

We know that, at t=0,

`T=148^(\circ)=Ae^0+C,`

solving the above equation for A, we get:

`A=148^(\circ)-67^(\circ)=81^(\circ).`

Now, setting T=102°, we get:

`102^(\circ)=81^(\circ)e^(-kt)+67^(\circ).`

Solving the above equation for t, we get:

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

Finally, we get:

`t\approx11.44.`

Answer:

`11.44`