a. What is the thaw rate of the turkey in hours per pound for refrigerator thawing? For cold water thawing? b. Write a linear function in the form y = mx + b to represent the time,... a. What is the thaw rate of the turkey in hours per pound for refrigerator thawing? For cold water thawing? b. Write a linear function in the form y = mx + b to represent the time, t, in hours it takes to thaw a turkey in the refrigerator as a function of the weight, w, in pounds of the turkey.
Understand the Problem
The problem presents two questions related to thawing a frozen turkey. Part (a) asks for the thaw rate in hours per pound for both refrigerator and cold water thawing methods. Part (b) requires the creation of a linear function, in slope-intercept form, to model the thawing time of a turkey in the refrigerator as a function of its weight.
Answer
a. Refrigerator: $6$ hours/pound, Cold Water: $0.5$ hours/pound b. $t = 6w$
Answer for screen readers
a. The thaw rate for refrigerator thawing is 6 hours per pound. The thaw rate for cold water thawing is 0.5 hours per pound. b. The linear function representing the thawing time in the refrigerator is $t = 6w$.
Steps to Solve
- Calculate the thaw rate for refrigerator thawing
The problem states that refrigerator thawing takes 1 day for every 4 pounds. Since we want the thaw rate in hours per pound, we need to convert 1 day into hours. There are 24 hours in a day.
$$ \frac{1 \text{ day}}{4 \text{ pounds}} = \frac{24 \text{ hours}}{4 \text{ pounds}} $$
$$ \frac{24 \text{ hours}}{4 \text{ pounds}} = 6 \text{ hours per pound} $$
- Calculate the thaw rate for cold water thawing
The problem states that cold water thawing takes 30 minutes per pound. We need to convert this to hours per pound. Since there are 60 minutes in an hour:
$$ \frac{30 \text{ minutes}}{1 \text{ pound}} = \frac{30 \text{ minutes}}{1 \text{ pound}} \cdot \frac{1 \text{ hour}}{60 \text{ minutes}} $$
$$ \frac{30 \text{ minutes}}{1 \text{ pound}} \cdot \frac{1 \text{ hour}}{60 \text{ minutes}} = \frac{30}{60} \text{ hours per pound} $$
$$ \frac{30}{60} \text{ hours per pound} = 0.5 \text{ hours per pound} $$
- Write the linear function for refrigerator thawing
We want to write a linear function in the form $y = mx + b$ where $y$ is the time, $t$, in hours, $x$ is the weight, $w$, in pounds, $m$ is the slope (thaw rate in hours per pound), and $b$ is the y-intercept (the time it takes to thaw a 0-pound turkey, which is 0). We already calculated the thaw rate for refrigerator thawing in step 1, which is 6 hours per pound.
$$ t = 6w + 0 $$
$$ t = 6w $$
a. The thaw rate for refrigerator thawing is 6 hours per pound. The thaw rate for cold water thawing is 0.5 hours per pound. b. The linear function representing the thawing time in the refrigerator is $t = 6w$.
More Information
Thawing food properly is very important to avoid illness!
Tips
A common mistake is not converting the units to be consistent (e.g., using days instead of hours or minutes instead of hours). Another common mistake is misunderstanding the relationship between rate, time, and weight, and incorrectly setting up the linear equation.
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