Respuesta :
Answer:
T = 76.39°C
Explanation:
given,
coffee cup temperature = 95°C
Room temperature= 20°C
expression
[tex]T( t ) = 20 + 75 e^{\dfrac{-t}{50}}[/tex]
temperature at t = 0
[tex]T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}[/tex]
T(0) = 95°C
temperature after half hour of cooling
[tex]T( t ) = 20 + 75 e^{\dfrac{-t}{50}}[/tex]
t = 30 minutes
[tex]T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}[/tex]
[tex]T( 30 ) = 20 + 75 \times 0.5488[/tex]
T(30) = 61.16° C
average of first half hour will be equal to
[tex]T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt[/tex]
[tex]T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30[/tex]
[tex]T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30[/tex]
[tex]T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750][/tex]
[tex]T = \dfrac{1}{30}[600 - 2058.04 + 3750][/tex]
T = 76.39°C
