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A pizza pan is removed at 7:00 PM from an oven whose temperature is fixed at 450°F into a room that is a constant 71°F. After
A pizza pan is removed at 7:00 PM from an oven whose temperature is fixed at 450°F into a room that is a constant 71°F. After
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Answer #1

Using Newton's law of cooling,

\small T(t)=T_{room}+(T_{0}-T_{room})e^{-kt}

T0 = initial temperature = 450oF

Troom = room/environment temperature = 71oF

k = cooling rate

After 5 minutes, T(t) = 300 oF

\small 300=71+(450-71)e^{-k\times 5}

\small \frac{300-71}{450-71}=e^{-k\times 5}

\small \frac{229}{379}=e^{-k\times 5}

\small 0.6=e^{-k\times 5}

\small ln(0.6)=ln(e^{-k\times 5})

\small -0.5108=-k\times 5

\small k=0.10216 min-1

a)

Temperature of pan = 130oF

\small 130=71+(450-71)e^{-0.10216t}

\small \frac{130-71}{450-71}=e^{-0.10216t}

\small \frac{59}{379}=e^{-0.10216t}

\small 0.15567=e^{-0.10216t}

\small ln(0.15567)=ln(e^{-0.10216t})

\small -1.86=-0.10216t

\small t=18

At 7:18 PM, temperature of pan is 130oF.

b)

T(t) = 230oF

\small 230=71+(450-71)e^{-0.10216t}

\small \frac{230-71}{450-71}=e^{-0.10216t}

\small \frac{159}{379}=e^{-0.10216t}

\small 0.42=e^{-0.10216t}

\small ln(0.42)=ln(e^{-0.10216t})

\small -0.8675=-0.10216t

\small t=8

The time that needs to elapse before the pan is 230oF is 8 minutes.

c)

As time tends to \small \infty , \small e^{-kt} tends to 1.

\small T(t)=T_{room}+(T_{0}-T_{room})=T_{0}

As time passes, temperature of pan is approaching 71oF.

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