Solving Battery Problem: Transport 2.6 kg Chocolate up 1000m

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SUMMARY

The discussion centers on calculating the mass of chocolate that can be transported using a 4.5 Volt, 1600 mAh battery in a lift scenario. The solution derived is 2.6 kg, based on the energy calculations using the formula J = C * V, where C is the charge in coulombs and V is the voltage. The participant correctly identifies the need to convert milliampere-hours to coulombs and calculates the total energy available from the battery as 25,920 Joules. The key question raised is how to relate this energy to lifting mass over a height of 1000 meters.

PREREQUISITES
  • Understanding of electrical energy calculations (Joules, volts, and ampere-hours)
  • Familiarity with gravitational potential energy equations
  • Basic knowledge of unit conversions (milliampere-hours to coulombs)
  • Ability to apply physics concepts to real-world scenarios
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  • Learn about gravitational potential energy and the formula PE = mgh
  • Research energy conversion efficiency in battery-operated systems
  • Explore the implications of battery capacity on transport systems
  • Investigate alternative energy sources for lifting applications
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Homework Statement



A 4.5 Volt is labeled: 4.5 V ; 1600 mAh (milliamperehour). This battery is the energy source of a chocolate lift. Find the number of kilograms of chocolater you can Transport with this lift up to the summit. The summit is 1000 meter above the valley Station.

Homework Equations



The solution is 2.6 kg

The Attempt at a Solution



I assume that my start is correct:

J = C * V

so

J = (1600 mAh / (3600 sec * 1000)) * 4.5 V

isn't it?

Is the solution 0.002 J then? What did I wrong?
How do I have to go on?

Thanks in advance!
 
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Now I found out my fault:

1 Ah = 3600 As = 3600 C
C = 1600/1000 * 3600 = 5760
V = 4.5

So:
J = 5760 * 4.5 = 25 920 J

Now my new question:
How can I find out the kilogram with help of the Joule?
 
You find a relevant equation for bringing a mass from the Valley Station to a summit that is 1000 m higher !

Ask yourself: how many Joules do I need to lift 1 kilogram over 1 m ?

Under "relevant equations" a potential helper expects to find the equations you think have to be brought into deal with the problem. "The answer is.." is not a relevant equation! Give me something better !
 

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