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A cake is removed from an oven at 232°F and left to cool at room temperature, which is 84°F. After 27 minutes, the temperature of the cake is 110°F. When will it be 85 F?

The cake will be at 85°F after _____ minutes.
(Round the final answer to the nearest whole number as needed. Round all intermediate values to six decimal places as needed.

1 Answer

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Final answer:

To determine when the cake will be 85°F, Newton's Law of Cooling is used to establish a relationship between the temperature of the cake and time. The constant rate of cooling is found by using the temperature after 27 minutes. Solving the equation with this constant will give the time in minutes at which the cake reaches 85°F.

Step-by-step explanation:

The cooling of the cake can be modeled using Newton's Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the ambient temperature. This can be represented by the formula T(t) = T_{​amb} + (T_{​init} - T_{​amb}) ​× e^{-kt}, where T(t) is the temperature of the object at time t, T_{​amb} is the ambient temperature, T_{​init} is the initial temperature of the object, and k is a constant that represents the cooling rate.




To find the constant k, we use the temperature of the cake after 27 minutes:

110 = 84 + (232 - 84) × e^{-27k}

Solving for k gives us the cooling rate.

Once k is known, we can solve for t when T(t) = 85°F by plugging in the values and solving the equation for t.

The solution will provide us the time in minutes when the cake will reach 85°F.

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