To answer the questions provided regarding the stove and heating of water, let's analyze each part:
(a) What value represents the useful work (or energy)?
The useful work (or energy) is the energy that is actually used for the intended purpose. In this case, it’s the energy gained by the pot of water, which is: Useful work = 1500 J
(b) What is the energy input?
The energy input is the total amount of energy supplied by the stove. According to the problem, this is: Energy input = 5000 J
(c) What is the energy output?
The energy output is the energy that is utilized by the system, which in this case is the energy gained by the pot of water plus any wasted energy (if any). However, here we consider only the energy actually used, which is: Energy output = 1500 J (the useful energy gained by the pot)
(d) If the pot of water is a system, is it open, closed or isolated? Explain your answer.
The pot of water can be considered an open system. This is because an open system allows for the exchange of matter and energy with its surroundings. In this case, as the stove heats the water, energy is being transferred to the system (the water), and vapor (water in gas form) can escape, indicating an exchange with the environment.
(e) Calculate the amount of wasted energy.
Wasted energy can be calculated by subtracting the useful work from the energy input: \[ \text{Wasted energy} = \text{Energy input} - \text{Useful work} = 5000 J - 1500 J = 3500 J \]
(f) Calculate the percent efficiency of the stove.
Percent efficiency can be calculated using the formula: \[ \text{Efficiency} = \left( \frac{\text{Useful work}}{\text{Energy input}} \right) \times 100 \] Plugging in the values: \[ \text{Efficiency} = \left( \frac{1500 J}{5000 J} \right) \times 100 = 30% \]
Summary of Answers:
(a) 1500 J
(b) 5000 J
(c) 1500 J
(d) Open system (because it exchanges energy and possibly matter with the environment)
(e) 3500 J
(f) 30%
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