Project 2

Given,

Battery resistance=0.5 mOhm

Maximum current= -0.896 ; Time,t= 120sec

Maximum power loss =I^2 *R = (-0.896)^2 *0.5 = 0.4mA

Joule heat = I^2 *R*t = (-0.896)^2 *0.5* 120sec = 0.048J = 48mJ

Max heat generated = joule heat + entropic heat = 48+2 = 50mJ