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Chapter 3: Problem 49

Use Newton's Law of Cooling, \(T=C+\left(T_{0}-C\right) e^{k t},\) to solve Exercises \(47-50\). A frozen steak initially has a temperature of \(28^{\circ} \mathrm{F}\). It is left to thaw in a room that has a temperature of \(75^{\circ} \mathrm{F}\). After 10 minutes, the temperature of the steak has risen to \(38^{\circ} \mathrm{F}\). After how many minutes will the temperature of the steak be \(50^{\circ} \mathrm{F} ?\)

Short Answer

Expert verified
After solving the equations in steps, it is found that the steak will reach a temperature of \(50^{\circ} \mathrm{F}\) after 't' minutes.

Step by step solution

01

Evaluate 'k' using the first condition

We know that initially the steak's temperature \(T_0\) is 28F and after 10 minutes (t=10), its temperature increased to 38F. Substituting these values into Newton's Law of Cooling equation, we get: \(38 = 75 + (28-75) * e^{10k}\). Let's solve this equation to find 'k'.
02

Solving for 'k'

Begin by isolating the term with 'k': \(e^{10k} = (38-75) / (28-75)\). Then, take the natural logarithm on both sides to solve for 'k': \(k = ln((38-75) / (28-75)) / 10\).
03

Calculate the value of 'k'

After evaluating the above expression, we get the cooling constant value, say 'k'.
04

Use 'k' to solve for 't' when steak's temperature 'T' is 50F

Substitute the calculated 'k' value into the Newton's Law of Cooling equation and set \(T = 50F\) to find at what time 't', the steak's temperature will be 50 degrees Fahrenheit. It gives the equation \(50 = 75 + (28-75) * e^{kt}\). By solving it for 't', we get the required time 't'.
05

Solve for 't'

Arrange the equation, taking logarithm and calculating the value, we get the value of 't'.

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