Question

An electric car accelerates for 8.0 s by drawing energy from its 320-V battery pack. During this time, 1300 C worth of electrons pass through the battery pack. (a) How many electrons are moved through the battery during this 8.0 s acceleration time? (b) How much energy transfer does this constitute? (c) Find the minimum horsepower rating of the car.(746 W = 1 hp)

Answer 1

(a) $8.13\cdot 10^{-21}$

The magnitude of the charge of one electron is

$q=1.6\cdot 10^{-19}C$

Here the total amount of charge that passed through the battery pack is

Q = 1300 C

So this total charge is given by

Q = Nq

where

N is the number of electrons that has moved through the battery

Solving for N,

$N=\frac{Q}{q}=\frac{1300 C}{1.6\cdot 10^{-19} C}=8.13\cdot 10^{-21}$

(b) $4.16\cdot 10^5 J$

First, we can find the current through the battery, which is given by the ratio between the total charge (Q = 1300 C) and the time interval (t = 8.0 s):

$I=\frac{Q}{t}=\frac{1300 C}{8.0 s}=162.5 A$

Now we can find the power, which is given by:

$P=VI$

where

V = 320 V is the voltage

I = 162.5 A is the current

Subsituting,

$P=(320 V)(162.5 A)=52,000 W$

And now we can find the total energy transferred, which is the product between the power and the time:

$E=Pt = (52,000 W)(8.0 s)=4.16\cdot 10^5 J$

(c) 69.7 hp

Now we have to convert the power from Watt to horsepower.

We know that

1 hp = 746 W

So we can set up the following proportion:

1 hp : 746 W = x : 52,000 W

And by solving for x, we find the power in horsepower:

$x=\frac{1 hp \cdot 52,000 W}{746 W}=69.7 hp$