what if? you are cooking aleecha, an ethiopian stew. when you take it off the stove, its temperature is…
what if? you are cooking aleecha, an ethiopian stew. when you take it off the stove, its temperature is 212°f. the room temperature is 70°f, and the cooling rate of the stew is r = 0.046. a. how long will it take to cool the stew to 100°f when the room temperature is 75°f? it will take about 37 minutes. b. in example 2, you found that you needed to wait about 34 minutes before the stew reached the 100°f serving temperature. how much sooner can your guests begin to eat when the serving temperature is 140°f instead of 100°f? about min sooner
Answer
Explanation:
Step1: Recall Newton's Law of Cooling formula
The formula for Newton's Law of Cooling is $T(t)=T_a+(T_0 - T_a)e^{-rt}$, where $T(t)$ is the temperature at time $t$, $T_a$ is the ambient (room) - temperature, $T_0$ is the initial temperature, $r$ is the cooling rate, and $t$ is the time.
Step2: First, find the time to cool to 140°F
We know that $T_0 = 212^{\circ}F$, $T_a = 70^{\circ}F$, $r = 0.046$, and $T(t)=140^{\circ}F$. Substitute these values into the formula: $140 = 70+(212 - 70)e^{-0.046t}$ $140-70=(212 - 70)e^{-0.046t}$ $70 = 142e^{-0.046t}$ $e^{-0.046t}=\frac{70}{142}\approx0.493$ Take the natural - logarithm of both sides: $\ln(e^{-0.046t})=\ln(0.493)$ $-0.046t=\ln(0.493)$ $t=\frac{\ln(0.493)}{-0.046}\approx\frac{- 0.707}{-0.046}\approx15.4$ minutes
Step3: Compare with the time to cool to 100°F
We know that it takes about 34 minutes to cool to 100°F. The difference in time is $34 - 15.4 = 18.6\approx19$ minutes
Answer:
19