You have been given Newton's Law of Cooling:

T(t) = Ts +(T0 - Ts)e^{-kt}

Here is an explanation of the variables:

t = time taken for the cooling

T(t) = the temperature of the given object at time t

Ts= the surrounding temperature

T0 = the initial temperature of the object

k = decay constant

Now that we understand what's what, let's see if we can solve the problem:

A pot of coffee with a temperature of 100 degrees Celsius is set down in a room with a temperature of 20 degrees Celsius. The coffee cools to 50 degrees celsius in one hour.

a. Find T(t), the temperature of the coffee at time t

b. How long will it take for the coffee to cool down?

a. Find T(t), the temperature of the coffee at time t

T(t) = Ts +(T0 - Ts)e^{-kt}

We don't have k. We can find it from the info given, however, since you know that in 1 hr, the temp went from 100 to 50:

t = time taken for the cooling == put in 1

T(t) = the temperature of the given object at time t == T(1) = 50

Ts= the surrounding temperature == we are given this as 20

T0 = the initial temperature of the object == given as 100

T(1) = 50 = 20 +(100 - 20)e^{-1k}

subtract 20 from both sides:

30 = (100 - 20)e-^{1k}

30 = (80)e^{-k}

divide by 80

0.375 = e^{-k}

take natural log of both sides:

ln(0.375) = ln(e^{-k})

use calc for left side, use power law for right side ln(ex) = x ln(e)

-0.980829 = -k ln(e)

ln(e) = 1, mult by (-1)

k = 0.980829 - I will round to 0.98

Now sub what we know into the general equation to get T(t) for this problem:

T(t) = 20 +(100 - 20)e^{-0.98t }

T(t) = 20 +80e^{-0.98t}

CHECK - at 1 hr, T = 50.02 -- the rounding I did with k affected it - but this is good enough.

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b. How long will it take for the coffee to cool down?

It is cool when the coffee reaches ambient temp, 20. So we want to put 20 in for T(t)

20 = 20 +80e^{-0.98t}

subtract 20 from both sides

0 =80e^{-0.98t}

divide by 80

0 = e^{-0.98t}

ln both sides

ln(0) = ln e^{-0.98t}

WAIT! Can't evaluate ln(0). That's because e, a positive number, to any power can never =0. The function has an asymptote there - goes on forever and never reaches 0, but approaches it. So we've got to find another way. Let's say we are happy if the temp is 20.5 - and see what happens --

20.5 = 20 +80e^{-0.98t}

0.5 =80e-0.98t

ln(0.5/80) = ln e-0.98t

[ln(0.5/80)]/(-0.98) = t = 5.2 hrs

using a table or a graph on a graphing calc we see at

t=6 hrs, T=20.2

t=8 hrs, T= 20.03

t=10 hrs, T=20.004

the difference is becoming insignificant. Maybe call it at 7.

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Understanding variables in equations is very important. Only then can we read a word problem and fill in the given info. Understanding directions is also very important. Finding T(t) means find the equation, and the only "letter" (variable) you can have in the resulting equation is the variable specified, t. In general, when you are given a set of conditions, you can use those to find unknowns. Part b was tricky and I took you the long way around, - to make sure you understood the math. From now on, if you have the equation, pump it into your graphing calculator and look at the graph to see what is going on.

All the best!