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Zepler [3.9K]
3 years ago
9

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, ho

w 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?
Physics
1 answer:
dexar [7]3 years ago
4 0

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

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