= Heat released to cold reservoir
= Heat released to hot reservoir
= maximum amount of work
= temperature of cold reservoir
= temperature of hot reservoir
we know that
eq-1
maximum work is given as
=
-
using eq-1
=
-
Answer:
Energy transferred = 28.8 Joules.
1. Energy transferred = 144 Joules.
2. The unit of potential difference, volts can also be described as Joules per Coulombs.
3. Current, I = 6.945 Amperes.
Explanation:
<u>Part A.</u>
Given the following data;
Current, I = 1.2A
Time, t = 2 minutes
Potential difference, V = 12 volts.
To find the energy transfered;
Energy transferred = charge moved * potential difference
E = Q * V
Substituting into the equation, we have;
Energy transferred = (1.2 * 2) * 12
Energy transferred = 2.4 * 12
Energy transferred = 28.8 Joules.
<u>Part B.</u>
1. <em><u>Given the following data;</u></em>
Charge, Q = 24C
Potential difference = 6V
To find the energy transferred;
E = Q * V
Substituting into the equation, we have;
E = 24 * 6
E = 144 Joules.
2. Since we know that, Energy transferred = charge moved * potential difference
The units of energy is Joules while the unit of the quantity of charge moved is Coulomb.
Therefore, the unit of potential difference becomes Joules per Coulomb.
3. <em><u>Given the following data;</u></em>
Potential difference = 18V
Energy transferred = 500J
Time, t = 4 minutes.
To find the current;
E = Q * V
Substituting into the equation, we have;
500 = Q*18
Q = 500/18
Q = 27.78C
But, Charge moved (Q) = current (I) * time (t)
Current, I = Q/t
Substituting into the equation, we have;
Current, I = 27.78/4
Current, I = 6.945 Amperes..