Final answer:
The heat expelled in one cycle of a two-cycle engine rated at 4500 Watts and 20 percent efficient, cycling 10 times a second, is 1800 Joules.
Step-by-step explanation:
To calculate the heat expelled per cycle by a two-cycle engine with a known power output and efficiency, we first need to understand that the efficiency of an engine (η) is given by the ratio of work output (M) to the heat input (Qh). Mathematically, this is expressed as η = W/Qh. Given that the engine is 20 percent efficient and operates at 4500 Watts, and it cycles 10 times a second, we can set up the following calculation.
Firstly, we need to determine the work done in one second by the engine, which can be gotten by multiplying the power output by 1 second because power is defined as work done per unit time (P = W/t). Thus, W = P × t = 4500 Watts × 1s = 4500 Joules.
Next, since the efficiency is 20%, meaning η = 0.20, we can calculate the heat input per second (Qh) using the efficiency formula η = W/Qh. Rearranging the equation, we get Qh = W/η. Then, Qh = 4500J/0.20 = 22500 Joules per second.
Given that the engine cycles 10 times per second, the heat input per cycle is Qh per cycle = Qh per second / number of cycles per second = 22500J / 10 = 2250 Joules per cycle.
Now, using the first law of thermodynamics, which states that the heat input (Qh) equals the work done (W) plus the heat expelled (Qc), we get Qc = Qh - W. Therefore, for one cycle, Qc per cycle = Qh per cycle - W per cycle = 2250J - 450J = 1800 Joules.
In conclusion, the heat expelled in one cycle by the two-cycle engine is 1800 Joules.