Question

A wire of resistance 5.0 Ω is connected to a battery whose emf is 2.0 V and whose internal resistance is 1.0 Ω. In 2.0 min, how much energy is (a) Transferred from chemical form in the battery, (b) Dissipated as thermal energy in the wire, and (c) Dissipated as thermal energy in the battery?

Short answer

1. The energy transferred from the chemical form in the battery is U=80J

2. The energy dissipated as thermal energy in the wire is U'=67J

3. The energy dissipated as the thermal energy in the battery is 13J

Step-by-step solution

Given

Resistance$R=5\Omega$

Emf$\epsilon =2V$

Internal resistance$r=1\Omega$

Time$t=2min$

Determining the concept

Use the formula of energy dissipation rate in terms of current and emf to find the energy transferred in the battery from the chemical energy and using the formula for the rate at which energy is dissipated in the form of current and resistance, find the energy dissipation as the thermal energy in the wire.

Formulae are as follow:

$$\begin{gathered} U=Pt \\ P=i\epsilon \\ {P}^{\text{'}}=i^{2}R \end{gathered}$$

Where,

P is power, t is time, i is current, 𝜀is emf, U is energy.

(a) Determining the energy transferred from the chemical form in the battery

The energy transferred from the chemical form in the battery:

Power is nothing but the energy per unit time and is given by,

$P=U/t$

The energy transferred is then,

U=Pt....................................1)

The rate P at which the chemical energy in the battery changes is,

$P=i\epsilon$

The current in a single loop circuit containing a single resistance R and an emf $\epsilon$device with single emfand internal resistance r is,

$i=\epsilon /(R+r)$

Hence,

$P=\frac{{\epsilon }^2}{(R+r)}$

Substituting this value in equation 1,

$U=\frac{\left({\epsilon }^{2}t\right)}{(R+r)}$

$U=\frac{\left(2V{)}^2\right(2min\left)\right(60s/min)}{(5\Omega +1\Omega )}$

U=80J

Hence, the energy transferred from the chemical form in the battery is 80J

(b) Determining the energy dissipated as thermal energy in the wire

The energy dissipated as thermal energy in the wire:

The rate at which energy is dissipated as the thermal energy in the wire is,

$P^{\text{'}}=i^{2}R=(\epsilon /(R+r){)}^{2}R$

Therefore, the energy dissipated as the thermal energy in the wire is,

$U^{\text{'}}=P^{\text{'}}t$$$\begin{gathered} U^{\text{'}}=Rt{\left(\frac{\epsilon }{(R+r)}\right)}^2 \\ {U}^{\text{'}}=(2V/(1\Omega +5\Omega \left){)}^2\right(5\Omega \left)\right(2min\left)\right(60s/min) \\ {U}^{\text{'}}=0.11111\times 600J \\ {U}^{\text{'}}=67J \end{gathered}$$

Hence, the energy dissipated as thermal energy in the wire is 67J.

(c) Determining the energy dissipated as the thermal energy in the battery

The energy dissipated as the thermal energy in the battery:

The difference U-U' gives the energy dissipated as the thermal energy in the battery.

$U-U^{\text{'}}=80J-67J=13J$

Hence, the energy dissipated as the thermal energy in the battery is 13J

Therefore, by using the formula of energy dissipation rate in terms of current and emf, the energy dissipated can be calculated.