Question

The temperature of a pan of hot water varies according to Newton's Law of Cooling: dT/ dt equals negative k times the quantity T minus A, where T is the water temperature, A is the room temperature, and k is a positive constant.
If the water cools from 90°C to 85°C in 1 minute at a room temperature of 30°C, how long (to the nearest minute) will it take the water to cool to 60°C?

a)3
b)4
c)5
d)8

Answer 1

@ganeshie8 @Luigi0210 @Destinymasha

Answer 2

@rational can u plz help? :)

Answer 3

@ParthKohli

Answer 4

@Ashleyisakitty can u plz help? :)

Answer 5

@nirmalnema

Answer 6

@satellite73

Answer 7

\[90-30=60\] and
\[85-30=55\] so you can model this as
\[T(t)=60\times \left(\frac{55}{60}\right)^t\] where\(T(t)\) is the difference in temperatures

Answer 8

or even
\[T(t)=60\times \left(\frac{11}{12}\right)^t\]

Answer 9

when the water is 60 degrees it will be \(60-30=30\) degrees hotter than the room
solve
\[30=60\times \left(\frac{11}{12}\right)^t\] using logs

Answer 10

so it would be:
1/2 = (11/12)^t
?:)

Answer 11

the solution is:
t=log(2)/log(11/12)

Answer 12

i got it, its 8 :) thank u!!!!!