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.