An engineer claims to have measured the characteristics of a heat engine that takes in 130J of thermal energy and produces 40J of useful work.
What is the smallest possible ratio of the temperatures (in kelvin) of the hot and cold reservoirs?
That would be whatever carnot cycle yields an efficiency of 4/13 = 30.8%
1 - T2/T1 = 0.308
T2/T1 = ?
TH/TC=(1)/(1-0.308)
To find the smallest possible ratio of the temperatures (in kelvin) of the hot and cold reservoirs for a heat engine, we can use the formula:
Efficiency = (Useful work output) / (Thermal energy input)
Given that the thermal energy input (Qh) is 130J and the useful work output (W) is 40J, we can calculate the efficiency:
Efficiency = (40J) / (130J)
Efficiency = 0.3077
Now, let's denote the temperature of the hot reservoir as Th and the temperature of the cold reservoir as Tc. According to the Carnot efficiency formula:
Efficiency = 1 - (Tc / Th)
We can rearrange the equation to solve for Tc / Th:
Tc / Th = 1 - Efficiency
Tc / Th = 1 - 0.3077
Tc / Th = 0.6923
To find the smallest possible ratio, we assume the value of Tc to be the smallest. Therefore, we have:
Tc / Th = 0.6923
Now, let's solve for Tc:
Tc = 0.6923 * Th
Hence, the smallest possible ratio of the temperatures (in kelvin) of the hot and cold reservoirs is 0.6923.
To find the smallest possible ratio of the temperatures, we can use the concept of the Carnot efficiency.
The Carnot efficiency is the maximum theoretical efficiency for a heat engine operating between two temperature reservoirs. It is given by the formula:
Efficiency = 1 - (Tc/Th)
where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.
In this case, the heat engine takes in 130J of thermal energy and produces 40J of useful work. The efficiency (η) of the heat engine is given by:
η = (W/ Qh)
where W is the work output and Qh is the heat input.
Substituting the given values, we have:
η = 40J / 130J
Simplifying, we get:
η = 0.3077
To find the smallest possible ratio of the temperatures, we need to find the maximum possible efficiency (Carnot efficiency). Rearranging the Carnot efficiency formula, we have:
Efficiency = 1 - (Tc/Th)
Rearranging the formula to solve for the ratio (Tc/Th), we get:
Tc/Th = 1 - Efficiency
Substituting the efficiency value, we have:
Tc/Th = 1 - 0.3077
Simplifying, we get:
Tc/Th = 0.6923
Therefore, the smallest possible ratio of the temperatures (Tc/Th) is 0.6923.
Note: This assumes that the heat engine operates on the Carnot cycle, which is an idealized process used for theoretical calculations. In real-world situations, the actual efficiency may be lower due to various factors like friction and heat losses.