Answered

Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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.)

Sagot :

pstnonsonjoku

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

Learn more about newtons law of cooling: https://brainly.com/question/14643865:

Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.