Question

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A cake recipe says to bake the cake until the center is 180 degrees F, then let the cake cool to 100 degrees F. The table shows temperature readings for the cake.

a. Given a room temperature of 69 degrees F, what is an exponential model for this data​ set?
b. How long does it take the cake to cool to the desired​ temperature?

Answer 1

Answer:

a. The exponential model is  $T(t)=69+(111)e^{-0.1368357021t}$  

b. It takes the cake 9.32 minutes to cool to the desired​ temperature

Step-by-step explanation:

Let us solve it by using the Newton's Law of cooling

$T(t)=C+(T_{0}-C)e^{kt}$ , where

• T(t) is the temperature at any given time
• C is the surrounding temperature

• $T_{0}$ is the initial temperature of the heated object
• k is a negative constant
• t is the time

∵ A cake recipe says to bake the cake until the center is 180 °F

∴ $T_{0}$ = 180 ⇒ initial temperature

∵ The room temperature is 69 °F

∴ C = 69 ⇒ surrounding temperature

- Use the table to substitute t and T to find the constant k

∵ At t = 5 minutes, T = 125 °F

∴ $125=69+(180-69)e^{5k}$

- Subtract 69 from both sides

∴ $56=(111)e^{5t}$

- Divide both sides by 111

∴ $\frac{56}{111}=e^{5k}$

- Insert ㏑ for both sides

∴ $ln(\frac{56}{111})=5k$

- Divide both sides by 5

∴ - 0.1368357021 = k

∴ $T(t)=69+(111)e^{-0.1368357021t}$  

a. The exponential model is $T(t)=69+(111)e^{-0.1368357021t}$  

∵ The cake cool to 100 °F

∴ The desired​ temperature is 100°

∴ T(t) = 100 ⇒ cake's temperature at t minute

- Use the model above to find t

∵ $T(t)=69+(111)e^{-0.1368357021t}$  

∴  $100=69+(111)e^{-0.1368357021t}$

- Subtract 69 from both sides

∴ $31=(111)e^{-0.1368357021t}$

- Divide both sides by 111

∴ $\frac{31}{111} =e^{-0.1368357021t}$

- Insert ㏑ for both sides

∴ $ln(\frac{31}{111})=-0.1368357021t$

- Divide both sides by - 0.136835702

∴ 9.321711938 = t

∴ t ≅ 9.32 minutes

b. It takes the cake 9.32 minutes to cool to the desired​ temperature