A dish of lasagna baked at 375°F is taken out of the oven into a kitchen that is 70°F. After 6 minutes, the temperature of the lasagna is 317.5°F. What will its temperature be 30 minutes after it was taken out of the oven?

To find the temperature of the lasagna 30 minutes after it was taken out of the oven, we will use the concept of heat transfer.

First, let's calculate the temperature change of the lasagna during the initial 6 minutes. The lasagna's temperature dropped from 375°F (oven temperature) to 317.5°F.

Temperature change during 6 minutes:
375°F - 317.5°F = 57.5°F

Now, let's determine the rate at which the lasagna cools down. We can use the formula for Newton's Law of Cooling:

dT/dt = -k(T - T_env)

where:
dT/dt is the rate of change of temperature with respect to time,
k is the cooling constant,
T is the temperature of the lasagna,
T_env is the ambient temperature (kitchen temperature).

Let's rearrange the formula to solve for k:

k = -1/(T - T_env) * dT/dt

Using the temperature change calculated earlier (57.5°F) and the time interval (6 minutes), we can calculate k:

k = -1/(375°F - 70°F) * 57.5°F/6 minutes

Simplifying the equation:

k = -1/305°F * 57.5°F/6 minutes

Now, we can use the calculated k to determine the temperature of the lasagna 30 minutes after it was taken out of the oven.

dT/dt = -k(T - T_env)

Let's plug in the given values:

dT/dt = -(-1/305°F * 57.5°F/6 minutes) * (T - 70°F)
dT/dt = 1/305°F * 57.5°F/6 minutes * (T - 70°F)
dT/dt = 0.019 out of °F/min * (T - 70°F)

Integrating the equation, we get:

∫(dT/(T - 70°F)) = ∫(0.019 out of °F/min) dt

Now, we can solve for T after the integration:

ln|T - 70°F| = 0.019 out of °F/min * t + C

where t is the time in minutes, and C is the constant of integration.

To solve for the constant C, we can use the initial condition T = 317.5°F at t = 6 minutes:

ln|317.5°F - 70°F| = 0.019 out of °F/min * 6 minutes + C

C = ln|247.5°F| - 0.114 out of °F/min

Now, we have the equation:

ln|T - 70°F| = 0.019 out of °F/min * t + ln|247.5°F| - 0.114 out of °F/min

To find the temperature T after 30 minutes, we substitute t = 30 minutes:

ln|T - 70°F| = 0.019 out of °F/min * 30 minutes + ln|247.5°F| - 0.114 out of °F/min

Now, we can solve for T:

ln|T - 70°F| = 0.57 out of °F + ln|247.5°F| - 0.114 out of °F/min

T - 70°F = e^(0.57 out of °F + ln|247.5°F| - 0.114 out of °F/min)

Finally, add 70°F to both sides to get the temperature of the lasagna 30 minutes after it was taken out of the oven.