Answer:
(a) the battery’s internal resistance is 0.4 Ω
(b) the dissipated power dissipated inside the battery is 40 W
(c) the rate 0.096°C/min
Explanation:
Given information:
emf, ε = 12 V
voltage, V = 16 V
current, I = 10 A
mass, m = 20 kg
specific heat, c = 0.300 kcal/kg°C = 1255.8 J/kg°C
(a) What is the battery’s internal resistance?
to calculate the internal resistance, we can calculate by using the following formula
V = ε - Ir
where
V = voltage (V)
I = current (A)
r = internal resistance (Ω)
ε = emf (V)
since the battery is being charged, the current is negative, so
V = ε - (-I) r
V = ε + Ir
r = (V- ε)/I
= (16 - 12)/10
= 0.4 Ω
(b) What power is dissipated inside the battery?
to determine the dissipated power in the battery, use the following equation
P = I²r
where
P = power (W)
P = (10)² (0.4)
= 40 W
(c) At what rate (in °C/min ) will its temperature increase if its mass is 20.0 kg and it has a specific heat of 0.300kcal/kg⋅°C , assuming no heat escapes
to find the rate of temperature increase by
Q = m c ΔT
where
Q = thermal energy (J)
c = specific heat (J/kg°C)
ΔT = the change of temperature
to find the energy, we use
E = Pt
the energy is converted in one minute
E = 40 x 60
= 2400 J
thus,
2400 = 20 x 1255.8 x ΔT
ΔT = 2400/(20 x 1255.8)
= 0.096°C/min
