A 5 $\Omega$ resistor is connected across a 6-Volt battery. Considering the energy dissipated as heat over a period of 10 seconds, and assuming constant resistance and voltage, det... A 5 $\Omega$ resistor is connected across a 6-Volt battery. Considering the energy dissipated as heat over a period of 10 seconds, and assuming constant resistance and voltage, determine the entropy generated in the resistor during this process, given an ambient temperature of 298 K.
Understand the Problem
The question asks us to calculate the entropy generated in a resistor when it's connected to a battery for a certain amount of time. We are given the resistance, voltage, time, and ambient temperature. To solve this, we'll first calculate the power dissipated by the resistor as heat. Then, we'll find the total heat generated over the given time period. Finally, we'll calculate the entropy generated by dividing the total heat by the ambient temperature.
Answer
$0.133 \, \text{J/K}$
Answer for screen readers
$0.133 , \text{J/K}$
Steps to Solve
- Calculate the power dissipated by the resistor
The power $P$ dissipated by a resistor is given by the formula:
$$ P = \frac{V^2}{R} $$
where $V$ is the voltage and $R$ is the resistance. Given $V = 10 , \text{V}$ and $R = 25 , \Omega$, we have:
$$ P = \frac{(10 , \text{V})^2}{25 , \Omega} = \frac{100}{25} = 4 , \text{W} $$
- Calculate the total heat generated
The total heat $Q$ generated over a time period $t$ is given by:
$$ Q = P \cdot t $$
where $P$ is the power and $t$ is the time. Given $P = 4 , \text{W}$ and $t = 10 , \text{s}$, we have:
$$ Q = 4 , \text{W} \cdot 10 , \text{s} = 40 , \text{J} $$
- Calculate the entropy generated
The entropy generated $\Delta S$ is given by:
$$ \Delta S = \frac{Q}{T} $$
where $Q$ is the heat and $T$ is the ambient temperature. Given $Q = 40 , \text{J}$ and $T = 300 , \text{K}$, we have:
$$ \Delta S = \frac{40 , \text{J}}{300 , \text{K}} = \frac{4}{30} = \frac{2}{15} \approx 0.133 , \text{J/K} $$
$0.133 , \text{J/K}$
More Information
The entropy generated represents the increase in disorder due to the heat dissipated in the resistor. This process is irreversible, as heat cannot spontaneously flow back into electrical energy in the resistor at a constant temperature.
Tips
A common mistake is forgetting to convert the units to SI units. In this problem, all units were already in SI units, so no conversion was required.
Another common mistake is using the wrong formula for power or entropy. Make sure to use the correct formulas: $P=V^2/R$ for power and $\Delta S = Q/T$ for entropy change in a thermal process.
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