Taking into account the "Second Law of thermodynaics", what is the maximum percentage of heat at 700 C that can be converted into electricity if the waste heat was produced as steam at 100 C?
Please Give as much detail as possible instructor will not help cause I missed this lecture because of a family emergency !! please
To find the maximum percentage of heat at 700°C that can be converted into electricity, we can use the concept of the Carnot efficiency based on the Second Law of Thermodynamics. The Carnot efficiency defines the maximum possible efficiency of a heat engine operating between two temperature reservoirs.
In this case, we have a waste heat source at 700°C and a heat sink at 100°C. The Carnot efficiency (ηc) is given by the formula:
ηc = 1 - (Tc/Th)
Where Tc is the temperature of the heat sink (converted to Kelvin) and Th is the temperature of the heat source (also converted to Kelvin).
So, first, we need to convert the temperatures from Celsius to Kelvin. To convert Celsius to Kelvin, we add 273.15 to the given temperature.
Tc = 100°C + 273.15 = 373.15 K
Th = 700°C + 273.15 = 973.15 K
Now we can calculate the Carnot efficiency:
ηc = 1 - (373.15 K / 973.15 K)
ηc ≈ 1 - 0.3831
ηc ≈ 0.6169 or 61.69%
Therefore, the maximum percentage of heat at 700°C that can be converted into electricity, if the waste heat is produced as steam at 100°C, is approximately 61.69%.