# A freshly brewed cup of coffee has temperature 95 degrees C in a 20 degree C room. When its...

## Question:

A freshly brewed cup of coffee has temperature 95{eq}^{\circ} {/eq}C in a 20{eq}^{\circ} {/eq}C room. When its temperature is 63{eq}^{\circ} {/eq}C, it is cooling at a rate of 1{eq}^{\circ} {/eq}C per minute.

When does this occur (in minutes)? (Round answer to two decimal places.)

## Newton's law of cooling

From Newton's law of cooling, it can be said that the heat loss rate is directly proportional to the difference in temperature of a given body and the surrounding in which it is placed.

## Answer and Explanation: 1

The equation for temperature of coffee using Newton's law of cooling :

{eq}T(t)=20+Ce^{kt} {/eq}

For initial time at t=0, the equation becomes

{eq}\\ 95=T(0)=20+Ce^{k\cdot 0}=20+C \\ C=75 {/eq}

So, the equation becomes,

{eq}T(t)=20+75e^{kt} {/eq}

After some time at {eq}t=t_o {/eq},

{eq}63=T(t_0)=20+75^{kt_0} \\75e^{kt_0}=43 \\e^{kt_0}=\frac{43}{75} = 0.573 \\kt_0=ln\ (\frac{43}{75})= -0.556 {/eq}

The rate of cooling is at a rate of 1{eq}^{\circ} {/eq}C per minute.

{eq}\\-1=T'(t_o)=75(e^{kt_o} k)=75ke^{kt_o} \\k=\frac{-1}{75e^{kt_o} }=\frac{-1}{75(0.573)}=-0.023 {/eq}

{eq}to=\frac{ln(\frac{43}{75})}{k}={\frac{-0.556}{-0.023}}=24.17\ minutes {/eq}

The temperature of coffee is 63{eq}^{\circ} {/eq}C after 24.17 minutes.

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