Thank you for visiting A lawnmower engine with an efficiency of 0 21 rejects 3600 J of heat every second What is the magnitude of the work that the. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Final answer:
To determine the work done by a lawnmower engine with a 0.21 efficiency that rejects 3600 J of heat every second, we can calculate it to be approximately 957 J of work per second.
Explanation:
When a lawnmower engine with an efficiency of 0.21 rejects 3600 J of heat every second, we need to calculate the work done by the engine in that same interval. To find the magnitude of the work, we use the formula efficiency (e) = work done (W) / total energy absorbed (Qh). The heat rejected (Qc) is essentially the difference between the total heat absorbed (Qh) and the work done (W), which can be expressed as Qc = Qh - W.
Since we know the efficiency and the heat rejected per second, we can set up the equation as follows:
0.21 = W / (W + 3600 J)
Solving for W, we multiply both sides by the denominator to get:
0.21W + 756 J = W
Subtracting 0.21W from both sides gives us:
756 J = 0.79W
W = 756 J / 0.79
W = 957 J (approximately)
Therefore, the engine does approximately 957 J of work every second.
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