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Answered

To cook fondant, sugar and water are heated to 230°F. Then the mixture
is cooled at room temperature before it is kneaded. If the cooked fondant
cools to 200°F in 15 minutes in a 68°F kitchen, how long will it take to
cool to 100°F? (Hint: Find the constant k first.)

Answer :

Pstnonsonjokugo

Using the Newton's law of cooling, the time it takes to cool to 100°F is 116 mins.

Newton's law of cooling

According to the Newton's law of cooling, the when an object is exposed to the surrounding it looses heat rapidly untill it is at a thermal equilibrium with the suurroundings.

Now;

T(t) = T(s) + (To - Ts)e^-kt

T(t) = temperature at time T

To = initial temperature

Ts = temperature of the surrounding

k = colling constant

t = time taken

200 = 68 + (230 - 68)e^-15k

200 = 68 + 162e^-15k

200 - 68/162 = e^-15k

ln(0.815) = -15k

k = ln(0.815)/-15

k = 0.014 min-1

Now;

100 = 68 + (230 - 68)e^-( 0.014t)

100 -68/162 = e^-( 0.014t)

0.198 =e^-( 0.014t)

ln(0.198) = -( 0.014t)

t = ln(0.198)/-0.014

t = 116 mins

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