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Average power consumption in watts

by enchanteuse
Tags: power
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enchanteuse
#1
Mar8-09, 06:50 PM
P: 10
1. The problem statement, all variables and given/known data

(a) If you follow a diet of 2140 food calories per day (2140 kilocalories), what is your average power consumption in watts? (A food or "large" calorie is a unit of energy equal to 4.2*10^3 J; a regular or "small" calorie is equal to 4.2 J.) Note for comparison that the power consumption in a table lamp is typically about 100 watts.
___ watts
(b) How many days of a diet of 2140 large calories are equivalent to the gravitational energy change from sea level to the top of Mount Everest, 8848 m above sea level? Assume your weight is 61 kg. (The body is not anywhere near 100% efficient in converting chemical energy into change in altitude. Also note that this is in addition to your basal metabolism.)
____ days

2. Relevant equations

Power = F * v
Power = W / delta t

I know that 1 watt = 1 J/s

3. The attempt at a solution

I converted 2140 calories to Joules, which is 2140 * (4.2*10^3) = 8988000 J
I'm not sure what I need to do from there because I can't figure out the work.

Any help would be appreciated!
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LowlyPion
#2
Mar8-09, 11:04 PM
HW Helper
P: 5,341
What is the definition of a watt?

If you have Joules in a day, and you can figure out how many seconds are in a day ...

As to Everest what is the potential energy increase required? m*g*h ?
enchanteuse
#3
Mar8-09, 11:10 PM
P: 10
I figured it out:)

a) (2140 cals *2400 J) / (24 days * 60 minutes * 60 seconds)

b) (61 kg * 8848 m * 9.8 m/s) / (2140 cals *2400 J)

LowlyPion
#4
Mar8-09, 11:40 PM
HW Helper
P: 5,341
Average power consumption in watts

Quote Quote by enchanteuse View Post
I figured it out:)

a) (2140 cals *2400 J) / (24 days * 60 minutes * 60 seconds)

b) (61 kg * 8848 m * 9.8 m/s) / (2140 cals *2400 J)
Weren't you using 4.2*10^3 J ?
enchanteuse
#5
Mar8-09, 11:42 PM
P: 10
Good catch! That was a typo...I meant 4.2 * 10^3 J.

Thanks LowlyPion:)


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