# A salad with a temperature of 43F is taken from the refrigerator and placed on the table in a room that is 68F. After 12 min the temperature of the salad is 55F. What will the temperature of the salad be after 20 min.

## well, you know that

dT/dt = k(68-T)

dT/(68-T) = k dt

ln(68-T) = kt + c1

68-T = e^(kt+c1)

68-T = ce^(kt)

T = 68 - ce^(kt)

We know that T(0) = 43, so c=25, and

T = 68-45e^(kt)

We know that T(12) = 55, so

55 = 68-45e^(12k)

45e^12k = 13

e^12k = (13/45)

12k = ln(13/45)

k = ln(13/45)/12 = -0.103

So, we have

T(t) = 68 - 45e^(-0.103t)

Now just find T(20)

## To determine the temperature of the salad after 20 minutes, we can use the concept of rate of change and assume that the salad's temperature follows a linear pattern.

We have two data points:

- Initial temperature of the salad (t=0): 43°F

- Temperature of the salad after 12 minutes (t=12): 55°F

From these two data points, we can calculate the rate of change (slope) of the salad's temperature. To do this, we use the formula:

Rate of change = (Change in temperature) / (Change in time)

Change in temperature = Final temperature - Initial temperature

= 55°F - 43°F

= 12°F

Change in time = Final time - Initial time

= 12 minutes - 0 minutes

= 12 minutes

Rate of change = 12°F / 12 minutes

= 1°F/minute

So, the salad's temperature increases by 1°F every minute.

To find the temperature after 20 minutes, we can multiply the rate of change by the number of minutes:

Temperature after 20 minutes = Initial temperature + (Rate of change * Time)

= 43°F + (1°F/minute * 20 minutes)

= 43°F + 20°F

= 63°F

Therefore, the temperature of the salad after 20 minutes will be 63°F.