Question

How much of the \(100 \mathrm{kJ}\) of thermal energy at \(650 \mathrm{K}\) can be converted to useful work? Assume the environment to be at \(25^{\circ} \mathrm{C}\).

Short answer

Answer: The maximum amount of useful work that can be extracted is 54.1 kJ.

Step-by-step solution

Identify the given information

We are given the following details: 1. Thermal energy available: \(Q_{in} = 100\,\text{kJ}\) 2. System's temperature: \(T_{hot} = 650\,\text{K}\) 3. Environment's temperature: \(T_{cold}=25^{\circ}\mathrm{C}\)

Convert Celsius to Kelvin

Since all the temperatures in thermodynamic calculations must be in Kelvin, we need to convert the environmental temperature from Celsius to Kelvin: \(T_{cold} = 25\,^{\circ}\mathrm{C} + 273.15 = 298.15\text{ K}\)

Calculate the Carnot efficiency

The Carnot efficiency of a heat engine is the maximum efficiency achievable in a system between two temperatures. It is given by the formula: \(\eta_{carnot} = 1 - \frac{T_{cold}}{T_{hot}}\) Substituting the given values: \(\eta_{carnot} = 1 - \frac{298.15\text{ K}}{650\text{ K}} = 0.541\)

Calculate the maximum work output

Now that we have the Carnot efficiency, we can find the maximum work output by multiplying the efficiency with the thermal energy available: \(W_{max} = \eta_{carnot} \times Q_{in}\) \(W_{max} = 0.541 \times 100 \,\text{kJ} = 54.1\,\text{kJ}\) So, the maximum amount of useful work that can be extracted from the 100 kJ of thermal energy at 650 K with an environmental temperature of 25°C is 54.1 kJ.