Question

Over a period of time, a hot object cools to the temperature of the surrounding air. This is described mathematically by Newton's Law of Cooling T = C+ (To-C) e-kt, where t is the time it takes for an object to cool from temperature Toto temperature T, C is the surrounding air temperature, and k is a positive constant that is associated with the cooling object. A cake removed from the oven has a temperature of 207°F and is left to cool in a room that has a temperature of 69°F. After 20 minutes, the temperature of the cake is 138°F. What is the temperature of the cake after 45 minutes? After 45 minutes, the temperature of the cake is ° (Do not round until the final answer. Then round to the nearest whole number as needed.)

Answer 1

1) In the first problem, it is given that over a period of time, a hot object cools to the temperature of the surrounding air and it is described mathematically by Newton's law of cooling , where 't ' is the time it takes for an object to cool from temperature 'T₀ ' to temperature 'T ', 'C ' is the surrounding air temperature, and 'k ' is a positive constant that is associated with the cooling object.

"T = C +(To-C) e-k -ht

Now, a cake removed from oven has a temperature of '207^oF' and is left to cool in a room that has a temperature of '69^oF'. After '20' minutes, the temperature of the cake is '138^oF'. From this data, we can find the positive constant 'k' that is associated with the cake.

We have, the initial temperature, T₀ = 207^oF

the final temperature, T = 138^oF

the surrounding air temperature, C = 69^oF

the time it took to reach the final temperature, t = 20 min.

Let's substitute all these values in the model equation of Newton's law of cooling.

=T=C+(To-C) e-kt

-2014 = 138 – 69 = (138) e

-20k

Applying 'natural log' function (ln = log_e) on both sides, we get

→loge (1) = loge (e-20k)

$\Rightarrow (-0.69315) = (-20k)$

-0.69315 -20 k

k = 0.0347

Now, the general equation for this cake scenario becomes .

Now, we have to find the temperature 'T ' of the cake after '45 ' minutes i.e., t = 45 min.

Substituting, T₀ = 207^oF , C = 69^oF & t = 45 min in the general equation, we get

→T=C +(To-C) -0.0317

$\Rightarrow T = 69 + (138)\;e^{-(1.5596)}$

→T = 69 + 29.0109

T= 98.0109 98°F

Therefore, after 45 minutes, the temperature of the cake is .

98°F"

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2)  In the second problem, it is given that the half-life of the radioactive element 'unobtanium-47' is '10 seconds'.

Initially, '176 ' grams of unobtanium-47 is present. We have to find the mass present after 10, 20, 30, 40 & 50 seconds respectively.

Usually, it can be done with the help of radioactive decay formula , where 'A₀ ' is the initial amount, 't ' is the time taken to reach the amount 'A ' and 'k ' is a constant.

"A= AC

But here, we are going to solve it in a very simple method since the required times of calculation are in series and related to the half-life time as in this problem.

Initially, the amount of unobtanium-47 present, m₀ = 176 gms.

Since it's half-life is 10 seconds, it is clear that after every 10 seconds, the amount present becomes half.

After 10 seconds:

we know that after 10 seconds, the amount becomes half i.e., m₁ = (m₀/2) = (176/2) = 88 gms.

After 20 seconds:

we know that again after 10 seconds (total 20 sec), the amount becomes half i.e., m₂ = (m₁/2) = (88/2) = 44 gms.

After 30 seconds:

we know that again after 10 seconds (total 30 sec), the amount becomes half i.e., m₃ = (m₂/2) = (44/2) = 22 gms.

After 40 seconds:

we know that again after 10 seconds (total 40 sec), the amount becomes half i.e., m₄ = (m₃/2) = (22/2) = 11 gms.

After 50 seconds:

we know that again after 10 seconds (total 50 sec), the amount becomes half i.e., m₅ = (m₄/2) = (11/2) = 5.5 gms.

Therefore,

The amount left after 10 seconds is grams.

88

The amount left after 20 seconds is $\mathbf{\underline{"44"}}$ grams.

The amount left after 30 seconds is grams.

The amount left after 40 seconds is grams.

The amount left after 50 seconds is grams.