Question

A wire of resistance 5.9 Ω is connected to a battery whose emf ε is 4.0 V and whose internal resistance is 1.2 Ω. In 2.9 min, how much energy is (a) transferred from chemical to electrical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated as thermal energy in the battery?

Answer 1

Answer:

a) 390J

b) 322J

c) 68J

Explanation:

We need to calculate the power given by the battery. the power is given by:

$P=V*I\\I=\frac{V}{R}\\I=\frac{4}{5.9+1.2}\\I=0.56A\\P=2.24W$

Watts is J/s so:

$E=P*t\\E=2.24\frac{J}{s}*2.9min*(60\frac{s}{min})=390J$

The thermal energy in the wire is given by:

$E_w=P_w*t\\P_w=I^2*R_w=0.56^2*5.9=1.85W\\E_w=1.85\frac{J}{s}*2.9min*(60\frac{s}{min})=322J$

And the the dissipated thermal energy in the battery will be the remainig energy:

$E_b=E-E_w\\E_b=390-322=68J$