Question

A 12.0-V emf automobile battery has a terminal voltage of 16.0 V when being charged by a current of 10.0 A. (a) What is the battery’s internal resistance? (b) What power is dissipated inside the battery? (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?

Answer 1

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