Answer:
After 21.28 minutes will the coffee be only lukewarm (30∘C)
Step-by-step explanation:
Given -
A travel mug of 87∘C coffee is left on the roof of a parked car on a cold winter day . The temperature of the coffee after t minutes is given by
[tex]H = 87(2)^{\frac{-t}{14}}[/tex]
Let after [tex]t_{1}[/tex] time H will be [tex]$30^\circ$[/tex]C
put t = [tex]t_{1}[/tex] , H = [tex]$30^\circ$[/tex]C
[tex]30 = 87(2)^{\frac{-t_{1}}{14}}[/tex]
[tex]\frac{30}{87} = (2)^{\frac{-t_{1}}{14}}[/tex]
.3448275 = [tex](2)^{\frac{-t_{1}}{14}}[/tex]
Taking logarithm both side
[ [tex]log(2^{x}) = xlog2[/tex] ]
log.3448275 = [tex]{\frac{-t_{1}}{14}}[/tex] log 2
-.4581 = [tex]{\frac{-t_{1}}{14}} \times.3010[/tex]
[tex]{\frac{-t_{1}}{14}}[/tex] = -1.52
[tex]t_{1}[/tex] = 21.28 minutes
