Question

Use Newton's Law of Cooling, T = C + (T0 – C).ekt, to solve the problem. A cup of coffee with temperature 102°F is placed in a freezer with temperature 0°F. After 8 minutes, the temperature of the coffee is 52.5°F. What will its temperature be 13 minutes after it is placed in the freezer? Round your answer to the nearest degree.

Answer 1

Answer

35°F

Step-by-step explanation:

$T= C + (T_0 - C)e^{kt}$

C is the temperature of environment = 0°F

T_0  is the initial temperature of object = 102°F

t is the time period = 8

Plug in all the values and solve for k

$T= 0 + (102 - 0)e^{k(8)}$

WE know T is  52.5°F after 8 minutes

solve for k

$52.5= 102e^{8k}$

Divide both sides by 102

$\frac{52.5}{102} = e^{8k}$

Take ln on both sides

$ln\frac{52.5}{102} = lne^{8k}$

$ln\frac{52.5}{102} = 8k$

Divide both sides by 8

k=-0.08302

$T= 0 + (102 - 0)e^{k(t)}$

Now we find out T  when t= 13

$T= 0 + (102 - 0)e^{-0.08302(13)}$

T= 34.66°F

So T= 35°F