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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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Chapter
Section
BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

On a typical day, a 65-kg man sleeps for 8.0 h, does light chores for 3.0 h. walks slowly for 1.0 h, and jogs at moderate pace for 0.5 h. What is the change in his internal energy for all these activities?

To determine
The change in internal energy for all his activities.

Explanation

Section1:

To determine: the change in internal energy for sleeping.

Answer: Change in internal energy for sleeping is 2.3MJ .

Explanation:

Given Info:

The weight of the man is 65.0kg .

Time of sleeping is 8h .

The metabolic rate for sleeping is 80J/s .

Formula to calculate the change in internal energy is,

ΔU=P(Δt)

  • P is the power or time rate of energy usage
  • Δt is the time taken for the activity

Substitute 80J/s for P and 8h for Δt to find the change in internal energy,

ΔUs=(80J/s)(8h)(3600s/h)=2.3×106J=2.3MJ

Thus, change in internal energy for sleeping is 2.3MJ .

Section2:

To determine: the change in internal energy for light chores.

Answer: Change in internal energy for light chores is 2.5MJ .

Explanation:

Given Info:

The weight of the man is 65.0kg .

Time of light chores is 3h .

The metabolic rate for light chores is 230J/s .

Formula to calculate the change in internal energy is,

ΔU=P(Δt)

  • P is the power or time rate of energy usage
  • Δt is the time taken for the activity

Substitute 230J/s for P and 3h for Δt to find the change in internal energy,

ΔUl=(230J/s)(3h)(3600s/h)=2.5×106J=2.5MJ

Thus, change in internal energy for light chores is 2.5MJ .

Section3:

To determine: the change in internal energy for walking slowly.

Answer: Change in internal energy for walking slowly is 8.3×105J .

Explanation:

Given Info:

The weight of the man is 65.0kg .

Time of walking slowly is 1h .

The metabolic rate for light chores is 230J/s .

Formula to calculate the change in internal energy is,

ΔU=P(Δt)

  • P is the power or time rate of energy usage
  • Δt is the time taken for the activity

Substitute 230J/s for P and 1h for Δt to find the change in internal energy,

ΔUw=(230J/s)(1h)(3600s/h)=8.3×105J

Thus, change in internal energy for light chores is 8

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