Average power consumption in watts

AI Thread Summary
The discussion focuses on calculating average power consumption based on a daily diet of 2140 kilocalories, converting this energy into joules to find the power in watts. Participants clarify the conversion process, noting that 1 watt equals 1 joule per second. The calculations involve determining the total joules consumed in a day and the time in seconds to derive the average power. Additionally, the discussion includes a comparison of energy required to ascend Mount Everest, factoring in weight and gravitational potential energy. Participants successfully resolve the calculations and correct any typographical errors in their energy conversion factors.
enchanteuse
Messages
10
Reaction score
0

Homework Statement



(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

Homework Equations



Power = F * v
Power = W / delta t

I know that 1 watt = 1 J/s

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!
 
Physics news on Phys.org
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 ?
 
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)
 
enchanteuse said:
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 ?
 
Good catch! That was a typo...I meant 4.2 * 10^3 J.

Thanks LowlyPion:)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top