Chem help please?

Question

The starting voltage of a fresh alkaline battery is 1.45 V The voltage decreases during discharge and is 0.70 V when the battery has delivered its rated capacity If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electrical work the battery could perform during discharge.

An “AA” alkaline battery yields a nominal capacity of 2850 mAh

Answer 1

work is a function of power

power is a function of voltage

the question says Voltage is linear so:

V(t)= at +V_starting (a is just a slope)

to find the slope: we need two data points : (time0, Vstarting) , (endtime, Vfinal)

so a= Vfinal-Vstarting / endtime-time0

=0.70- 1.45 / 3600s ( nominal capacity of 2850 mAh--> 1hr-->3600s)

(this is the slope, calculate it)

so V(t) = at +1.45

P(t) = V*I = V(t) *I (I= 2850 mA)

then W= ∫₀³⁶⁰⁰ P(t) dt

= ∫₀³⁶⁰⁰ ( at +1.45 )*I dt

= I* ∫₀³⁶⁰⁰ ( at +1.45 )dt

you now have everything, you just need to do the integral

I= 2850 mA = 2.850 A

integral = 2.850* (at²/2 + 1.45t) ]₀³⁶⁰⁰

so W= 2.850( a*3600² /2 + 1.45*3600)

= this is your answer!