Answer:
(a) the temperature of the drink after 45 minutes is 12.77[tex]^{o}C[/tex]
(b) its temperature be 14[tex]^{o}C[/tex] at t = 56.50 Minutes
Step-by-step explanation:
Given information:
Initial temperature, [tex]T_{0}[/tex] = 5[tex]^{o}C[/tex]
Sorrounding temperature, C = 20[tex]^{o}C[/tex]
in 25 minute, T(25) = 10[tex]^{o}C[/tex]
(a) the temperature of the drink after 45 minutes
According Newton's law of cooling
T(t) = C + ([tex]T_{0}[/tex] - C)[tex]e^{-kt}[/tex]
where
[tex]T_{0}[/tex] = initial temperature
t = time
k = cooling constant
C = sorrounding temperature
Thus,
T(25) = 20 + (5 - 20)[tex]e^{-25k}[/tex]
10 = 20 - (15 [tex]e^{-25k}[/tex])
[tex]e^{-25k}[/tex] = [tex]\frac{20-10}{15}[/tex]
-25k = ln([tex]\frac{20-10}{15}[/tex])
k = - (ln([tex]\frac{20-10}{15}[/tex]))/25
= 0.016
hence, the complete equation
T(t) = 20 + (5 - 20)[tex]e^{-0.016t}[/tex]
in 45 minute the temperature is
T(45) = 20 + (5 - 20)[tex]e^{-0.016 x 45}[/tex]
T(45) = 12.77[tex]^{o}C[/tex]
the temperature is 14[tex]^{o}C[/tex], so T(t) = 14[tex]^{o}C[/tex]
T(t) = 20 + (5 - 20)[tex]e^{-0.016t}[/tex]
14 = 20 + (5 - 20)[tex]e^{-0.016t}[/tex]
[tex]e^{-0.016t}[/tex] = [tex]\frac{20-14}{15}[/tex]
-0.016t = ln ([tex]\frac{20-14}{15}[/tex])
t = - (ln ([tex]\frac{20-14}{15}[/tex]))/ 0.016)
t = 56.50 Minutes
