A heavy-duty AA alkaline dry cell\'s (+) terminal is 1.6 [V] higher than its ()

Question

A heavy-duty AA alkaline dry cell's (+) terminal is 1.6 [V] higher than its () terminal; the (+) end can deliver 12000 C from a slow chemical reaction.
a) How much electric Potential Energy does it start with?
b) What Power production (in [Watt] !) will drain that battery in 50 hours?
At this drain rate, the "terminal voltage" starts at 1.54[V] , decreasing to 1.28[V] after 25 hour and 1.02[V] after 50 hour.
c) How much Energy does the user get from the battery? . . . how efficient is that discharge? (slower discharge => terminal closer to 1.6V)

Explanation / Answer

a). U = qV = 12000 x 1.6 = 19.2KJ

b). P = U/t = 19200/(50 x 60 x 60) = 0.107 Watt

c) User voltage: 1.54 to 1.28 to 1.04

So, Vavg = 1.28V

Euser = qVuser = 12000 x 1.28 = 15.36KJ

d). e= Euser/ U = 15.36/19.2 = 0.80 = 80%

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