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A pound of body fat stores an amount of chemical energy equivalent to 3500 Cal. When sleeping, the average adult burns or expends about 0.45 Cal/h for every pound of body weight. How many Calories would a 122 lb person burn during 7 hours of sleep?

Answer: ____ Cal

Answer :

Final answer:

A 122 lb person would burn approximately 38.61 Calories during 7 hours of sleep, calculated by multiplying the Calories burned per hour per pound by the person's weight and the number of hours slept.

Explanation:

To calculate how many Calories a 122 lb person burns during 7 hours of sleep, we can use the given burn rate of 0.45 Cal/h per pound of body weight. We perform the following calculation:

Total Calories burned = (Calories burned per hour per pound) x (body weight in pounds) x (number of hours of sleep)

Total Calories burned = 0.45 Cal/h/lb x 122 lbs x 7 h = 38.61 Cal

Therefore, a 122 lb person would burn approximately 38.61 Calories during 7 hours of sleep.

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Rewritten by : Barada

Given:

• Amount of fat stored = 3500 Cal

,

• Average rate of fat burned = 0.45 Cal/h

,

• Weight of person = 122 lb

,

• Time = 7 hours

Let's find the number of calories the person will burn.

Since the average rate an adult burns is 0.45 Cal/h for every pound, it means the amount the person will burn per hour is:

A = 0.45 x 122 = 54.9 Cal/h

Now, the person slept for a total of 7 hours.

The total amount the person will burn for 7 hours will be:

54.9 x 7 = 384.3 Cal.

Therefore, the amount of calories the person will burn is 384.3 Cal.

ANSWER:

384.3 Cal.