Question

You are cooking beef stew. When you take the beef stew off the stove, it has a temperature of 2008F. The room temperature is 758F and the cooling rate of the beef stew is r 5 0.054. How long (in minutes) will it take to cool the beef stew to a serving temperature of 1008F

Answer 1

Answer:

Therefore it would take about 30 minutes to cool the beef stew to a serving temperature of 1008F

Explanation:

Given that:

The initial temperature of the stew(I) is 2008° F

The room temperature(R) is 758°F

The cooling rate of the beef stew(r) = 0.054

you need the beef stew to cool to a serving temperature(F) of 1008°F

To calculate How long (in minutes) will it take to cool the beef stew to a serving temperature of 1008°F we use the equation:

$F=(I-R)e^{-rt} +R$

Substituting values:

$1008=(2008-758)e^{-0.054t} +758$

$1008=1250e^{-0.054t} +758$

$1008-758=1250e^{-0.054t}$

$250=1250e^{-0.054t}$

$\frac{250}{1250}=\frac{1250}{1250}e^{-0.054t}$

$0.2=e^{-0.054t}$

taking natural log of both side

$ln(0.2)=ln(e^{-0.054t})$

$-1.6094=-0.054t$

$t=29.8$

Therefore it would take about 30 minutes to cool the beef stew to a serving temperature of 1008F