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
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
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
To determine the rise in temperature of the battery during 1 minute of cranking the engine, we can use the formula:
ΔT = (I^2 * R * t) / (m * c)
Where:
ΔT is the change in temperature
I is the current (in Amperes)
R is the internal resistance (in Ohms)
t is the time (in seconds)
m is the mass of the battery (in kilograms)
c is the specific heat capacity (in kcal/kg K)
First, let's convert the time from minutes to seconds since the other values are in SI units. In 1 minute, there are 60 seconds.
t = 1 minute * 60 seconds/minute
t = 60 seconds
Now let's substitute the given values into the formula:
ΔT = (80 A^2 * 12 Ω * 60 s) / (20 kg * 0.2 kcal/kg K)
Calculating further:
ΔT = (57600 A^2 * s * Ω) / (4 kcal K/kg)
ΔT = 14400 A^2 * s * Ω / kcal K/kg
Since 1 kcal = 4.184 kJ, we can convert kcal to kJ:
ΔT = 14400 A^2 * s * Ω / (4.184 kJ K/kg)
ΔT = 3446.823 kg A^2 * s * Ω / kJ K
Therefore, the rise in temperature of the battery during 1 minute of cranking the engine is approximately 3446.823 kg A^2 * s * Ω / kJ K.