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

Question

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

i get 178.89 to two decimal places

amistre64 · Answer

should 133 be 203?

anonymous · Answer

no

amistre64 · Answer

i see it now lol   203 - 70 right?

anonymous · Answer

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

anonymous · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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