An outboard motor for a boat is cooled by lake water at 16.0°C and has a compression ratio of 13.3. Assume that the air is a diatomic gas.
a) Calculate the efficiency of the engine's Otto cycle.
b) Using your answer to part (a) and the fact that the efficiency of the Carnot cycle is greater than that of the Otto cycle, calculate the maximum temperature (in °C) of the engine.
To calculate the efficiency of the engine's Otto cycle, we need to use the equation:
Efficiency = 1 - (1 / compression ratio)^(γ - 1)
where γ is the specific heat ratio for a diatomic gas, which is typically around 1.4.
a) To calculate the efficiency, we can substitute the given values into the equation:
Compression ratio = 13.3
Specific heat ratio (γ) = 1.4
Efficiency = 1 - (1 / 13.3)^(1.4 - 1)
Now we can use a calculator to evaluate this expression:
Efficiency = 1 - (1 / 13.3)^0.4
Efficiency ≈ 0.537 or 53.7%
Therefore, the efficiency of the engine's Otto cycle is approximately 53.7%.
b) To find the maximum temperature of the engine using the efficiency, we can use the formula for the efficiency of a Carnot cycle:
Efficiency of a Carnot cycle = 1 - (T_cold / T_hot)
Given that the efficiency of the Carnot cycle is greater than that of the Otto cycle, we can set the efficiency of the Carnot cycle equal to the efficiency of the Otto cycle:
1 - (T_cold / T_hot) = 1 - 0.537
Simplifying this equation:
0.537 = T_cold / T_hot
Here, T_cold is the temperature of the lake water at 16.0°C, which we need to convert to Kelvin:
T_cold = 16.0 + 273.15 = 289.15 K
Now we can solve for T_hot, the maximum temperature of the engine:
0.537 = 289.15 K / T_hot
Simplifying further:
T_hot = 289.15 K / 0.537
Now we can use a calculator to evaluate this expression:
T_hot ≈ 538.94 K
Finally, we need to convert the temperature back to Celsius:
T_hot ≈ 538.94 - 273.15 ≈ 265.79°C
Therefore, the maximum temperature of the engine is approximately 265.79°C.