Electric Potential Homework: Calculate Water Mass Heated to 100 ˚C

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SUMMARY

The discussion focuses on calculating the mass of water that can be heated to 100 ˚C using the work done by an electric force due to a potential difference of 1.2 × 10^9 V and a charge of -35 C. The formula derived is m = -q (Vcloud - Vground) / [gamma] * difference in temp, where [gamma] represents the heat capacity per unit mass of water. The conversion between calories and Joules is emphasized, clarifying that one calorie heats one gram of water by one degree Celsius.

PREREQUISITES
  • Understanding of electric potential difference and its implications
  • Knowledge of the relationship between charge, voltage, and work
  • Familiarity with heat capacity and its role in thermal energy calculations
  • Basic concepts of energy units, specifically calories and Joules
NEXT STEPS
  • Research the specific heat capacity of water and its implications in thermal calculations
  • Learn about the conversion factors between calories and Joules in energy calculations
  • Study the principles of electric potential and work done in electrostatics
  • Explore the relationship between temperature change and energy transfer in thermodynamics
USEFUL FOR

This discussion is beneficial for physics students, electrical engineers, and anyone involved in thermodynamics or energy conversion calculations.

arod2812
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Homework Statement


Suppose a potential difference of Vcloud - Vground = 1.2 × 10^9 V exists between the cloud and the ground, and q = -35 C of charge is transferred from the ground to the cloud.

If the work done by the electric force were converted into heat, what mass m of water at 0 ˚C could be heated to 100 ˚C?


Homework Equations


what is gamma in this equation? Or is it even gamma?

Is the difference in temperature 100˚C?

The Attempt at a Solution


m= -q (Vcloud - Vground) / [symbol for gamma]* difference in temp.

= -(-35 C)(1.2x10^9 V) / [gamma]* 100 ˚C
 
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It may help to recall that one calorie is the heat needed to heat one cc = 1gram of water by one degree C.

Calories are units of energy distinct from the Joules you get when you multiply Coulombs by volts.

Your gamma appears to be the heat capacity per unit mass of water but that you'll see is also just the unit conversion factor between Joules and calories.
 

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