a certain 12v storage battery with an internal resistance of 12 ohms delivers 80 A when used to crank a gasoline engine. if the battery mass is 20 kg and it has average specific heat of 0.2 kcal/kg K. what is its rise in temperaure during 1 min of cranking the engine. assume 100% efficiency

Answer 1

Power = i^2 R
energy = power * time = i^2 R t in Joules
t = 60 seconds
set that equal to heat energy used to warm the battery
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0.2 kcal/kg K = Joules/kg K
to get C = 836.8 J/kg K

i^2 R t = 836.8 * 20kg * (delta T)
solve for delta T

Answer 2

Assuming 100% efficient, there must be a zero internal resistance.
P=VI
(12V)(80A)
=960 W

E=Pt
Let E be Q which is the heat energy
t=60s
Q=Pt
=(960)(60)
=57600 J

To solve for delta T, we use the equation:

delta T=Q/cm
covert 0.2kcal/kg K to J/kg K
so the answer is 837.2 J/kg K

substituting Q, c and m to formula,
delta T= 57 600/837.2J/kgK(20)
So delta T= 3.44 K