Question

65. A pizza, heated to a temperature of 400°F, is taken out of an oven and placed in a 75°F room at timet= 0 minutes. The temperature of the pizza is changing such that its decay constant, k, is 0.325. At what time is the temperature of the pizza 150°F and, therefore, safe to eat? Give your answer in minutes.

Answer 1

The temperature of the pizza will be 150°F and safe to eat after approximately 10.166 minutes.

Step-by-step explanation:

To find the time at which the temperature of the pizza is 150°F and safe to eat, we can use the formula for exponential decay: T(t) = T0 * e^(-kt), where T(t) is the temperature at time t, T0 is the initial temperature, k is the decay constant, and e is the base of the natural logarithm.

Substituting the given values, we have T(t) = 400 * e^(-0.325t).

To find the time when the temperature is 150°F, we set T(t) equal to 150 and solve for t: 150 = 400 * e^(-0.325t).

Dividing both sides by 400 gives us e^(-0.325t) = 0.375. Taking the natural logarithm of both sides, we have -0.325t = ln(0.375).

Finally, we solve for t: t = ln(0.375) / -0.325 ≈ 10.166 minutes.

Answer 2

769MIN

Step-by-step explanation:

Hello!

To solve this problem we must use the straight line equation, this equation is as follows

Y=mx+b

where

X= time

Y=Temperature

b= intercept with the y axis when x = 0, therefore y = 400

m=k= temperature decrease with respect to time=-0.325

now we replace the values ​​in the equation of the line

Y=-0.325X+400

Now we find the time it takes for the pizza to reach 150 F replacing in the equation and finding X

150=-0.325X+400

150-400=-0.325X+

`X=(150-400)/(-0.325) =769MIN`