Percent change/exponential functions

Question

A cup of coffee at 100°C cooled to 46.3°C after 15 min

determine the percent decrease of temperature in each minute

use an exponential function to predict the temperature of the coffee after 30 min

if another cup is cooled at the same rate to 36°C after 15 min, what’s the initial temperature?

please show each step, I don’t know how to do this

Answer 1

y=ab^x

46.3=100b^x

46.3/100=b^(-15)

0.463=b^(-15)

log(0.463)=log(b^[-15])

-0.334419=-15log(b)

-0.334419/(-15)=log(b)

0.0222946=log(b)

10^(0.0222946)=b

1.052675=b

y=100(1.052675)^(-30)

y=100(1/[4.664759])

y=21.437°

36=a(1.052675)^(-15)

36=a(1/[2.159805])

36*2.159805=a

a=77.75298°

Answer 2

set up the know relationship: 46.3 = 100 e^(-tk) where in this case t is 15 minutes. This allows us to solve for the constant k which defines the characteristic of this material and schooling factor. So solving the equation: 46.3 = 100 e^(-15*k), gives AK equal to0.0513.

Now repeat the calculation, this time using time equal to 30. This gives a result of 21.45 degrees (in 30 minutes). Assuming conditions being similar we can now solve for the initial temperature for a cup which is called to 36° and 15 minutes

So: 36=T(init) e^(-15*.0513)

36/T(init) = e^(-15*.0513) = .4632. Divide 36 by.4632 and you get 77.7° as the initial temperature