Work done by electric force heats water from 0°C to 100°C

[arod2812]

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

 

[jambaugh]

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.