### BRAVO! That is correct.

A cup of hot tea, initially at 95oC, cools to 75oC, in 5.0 minutes when sitting in 25oC sirrpimdomgs. Use Newton's Law of Cooling to determine how long it will take to cool to 35oC.

Newton's Law of Cooling is

Tsurf is Tsurf(t), the temperature of the surface as a function of time t.

To is the temperature of the surroundings, To = 25oC .

T is the initial temperature difference; T = 95oC - 25oC = 70 Co.

We know Tsurf at time t = 5.0 min; that is 75oC.

Tsurf = To + T e - t/

75oC = 25oC + (70oC)e - 5 min/

We know everything in this equation except the time constant . This describes the cooling rate and is determined by the physical situation. (Solving for gives opportunity to play with the natural logrithm and exponential functions on a calculator).

(70oC)e - 5 min/ = 75oC - 25oC = 50oC

e - 5 min/ = 50oC/70oC

e - 5 min/ = 0.714

How can we get out of this exponent? Recall -- from a long-ago algebra class -- that ln(ex) = x. That means that

ln(e - 5 min/ ) = - 5 min/

So we take the ln() -- the "natural logrithm" function -- of both sides of the equation,

ln(0.714) = - 0.3368

- 5 min/ = - 0.3368

5 min/ = 0.3368

/5 min = 1/0.3368

/5 min = 2.967

= (2.967)(5 min )

= 14.84 min

This is the time constant for the exponent in Newton's Law of Cooling. Now we can evaluate Tsurf for any time t. Or we can calculate the time t for any temperature Tsurf. What time t gives Tsurf = 35oC?

Tsurf = To + T e - t/

35oC = 25oC + (70oC)e - t/14.84 min

35oC - 25oC = (70oC)e - t/14.84 min

10oC = (70oC)e - t/14.84 min

(70oC)e - t/14.84 min = 10oC

e - t/14.84 min = 10oC/70oC

e - t/14.84 min = 0.143

And now -- just as before -- we need to use the ln() -- the "natural logrithm" function -- to get the time t out of the exponent,

ln(e - t/14.84 min) = ln( 0.143)

- t/14.84 min = - 1.95

- t/14.84 min = - 1.95

- t/14.84 min = - 1.95

t/14.84 min = 1.95

t = (1.95)(14.84 min)

t = 28.9 min