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A cake is removed from a [tex]350^\circ F[/tex] oven and placed on a cooling rack in a [tex]70^\circ F[/tex] room. After 30 minutes, the cake is [tex]200^\circ F[/tex]. When will it be [tex]100^\circ F[/tex]?

Answer :

Final answer:

The cake's temperature decreases according to Newton's Law of Cooling, and we can use this law to calculate the time it takes for the cake to reach a certain temperature.

Explanation:

The cake is removed from a 350°F oven and placed in a 70°F room. After 30 minutes, the cake's temperature decreases to 200°F. To find out when it will be 100°F, we can use Newton's Law of Cooling.

According to Newton's Law of Cooling, the rate of change of the temperature of an object is proportional to the difference between the object's temperature and the temperature of its surroundings. This can be expressed as:

T' = -k(T - Ts)

where T' is the rate of change of temperature with respect to time, T is the temperature of the object, Ts is the temperature of the surroundings, and k is the cooling constant. In this case, we can rearrange the equation to solve for time:

t = (1/k) * ln((T - Ts) / (T0 - Ts))

where t is the time, T0 is the initial temperature of the object, and ln is the natural logarithm.

Plugging in the values from the problem:

t = (1/k) * ln((200 - 70) / (350 - 70))

We can find the value of the cooling constant, k, by using the given information. Since we know the temperature dropped from 350°F to 200°F in 30 minutes, we can use this to find k:

-k = (T' / (T - Ts)) = (200 - 70) / (350 - 70) / 30

simplifying, we get:

k = -((200 - 70) / (350 - 70)) / 30

Now we can substitute the value of k into the equation for time:

t = (1 / -((200 - 70) / (350 - 70)))) * ln((200 - 70) / (350 - 70))

Calculating this equation will give us the approximate time it takes for the cake to reach 100°F.

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