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29 November, 13:53

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?

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  1. 29 November, 15:29
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    (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
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