## anonymous 5 years ago If a cup of tea has an initial temperature of 203 degree F in a room whose temperature is 70 degreeF, then according to Newton's law of cooling the temperature of the tea after t minute have passed is

1. anonymous

T (t) =70+133e^(-t/5) what is the average temperature of the tea one hour after the initial time?? please help!

2. anonymous

replace t by 1. $T(1)=70+133e^{\frac{-1}{5}}$

3. anonymous

i get 178.89 to two decimal places

4. amistre64

should 133 be 203?

5. anonymous

no

6. amistre64

i see it now lol 203 - 70 right?

7. anonymous

the room temp is 70, 203-70=133 and it is the difference that decays

8. anonymous

how do you get that answer of 178.89 from that equation? b/c i got something different :(

9. amistre64

yeah, so t = -60 minutes if that is accurate....

10. amistre64

70 + (133 * (e^((-60) / 5))) = 70.0008172 is what google figures out

11. amistre64

or is this a recursion equation.... and not a particular solution?

12. amistre64

nah; particular it is; no place for a T{n-1} lol