A chip dissipates 2 kJ of electric work and rejects it as heat transfer from its surface which is at 50°C to 25°C air. How much entropy is generated outside the chip?

Question

Quizplus · Accepted Answer

The entropy generated outside the chip can be calculated using the formula for entropy change, $\Delta S = \frac{Q}{T}$, where $Q$ is the heat transfer and $T$ is the temperature in Kelvin. The heat rejected by the chip is 2 kJ or 2000 J, and the air temperature is 25°C, which is 298 K in Kelvin. The entropy generated is thus $\Delta S = \frac{2000}{298} = 6.711 j/k$. However, the given options seem to be incorrect based on the calculation, suggesting a possible misunderstanding in the question's context or a mistake in the provided options. The correct approach to find the entropy generated involves using the heat transfer and the temperature, but none of the provided options match the calculated value.

A Chip Dissipates 2 KJ of Electric Work and Rejects