This is the content of the pop-over!





Praxis Physical Science (5485) Practice Tests & Test Prep by Exam Edge - Free Test


Our free Praxis Physical Science (5485) Practice Test was created by experienced educators who designed them to align with the official Educational Testing Service content guidelines. They were built to accurately mirror the real exam's structure, coverage of topics, difficulty, and types of questions.

Upon completing your free practice test, it will be instantly reviewed to give you an idea of your score and potential performance on the actual test. Carefully study your feedback to each question to assess whether your responses were correct or incorrect. This is an effective way to highlight your strengths and weaknesses across different content areas, guiding you on where to concentrate your study efforts for improvement on future tests. Our detailed explanations will provide the information you need to enhance your understanding of the exam content and help you build your knowledge base leading you to better test results.

Login or Create an Account to take a free test

After you have completed your free test you will receive a special promo code that will save your between 10-15% on any additional practice tests!


** Sample images, content may not apply to your exam **


Additional test information
Back To General Exam Info

Praxis Physical Science - Free Test Sample Questions

A coal-fired electric power plant burns coal at a rate of 100 tons per hour.  The rate of air flow into the combustion chamber is 1.3 million pounds per hour. The useful energy output from that plant is 37% of the input energy.  What must the mass rate of flue gases being exhausted from the plant be if the rate of ash production is 10 tons per hour?





Correct Answer:
1.48 million pounds per hour.
to determine the mass rate of flue gases being exhausted from the coal-fired power plant, we need to apply the principle of conservation of mass. this principle states that the total mass entering a system must equal the total mass exiting the system when the system is at a steady state. we can use this principle to set up an equation accounting for all inputs and outputs of the system.

let's break down the inputs and outputs: 1. **inputs**: - coal is being burned at a rate of 100 tons per hour. since 1 ton is equal to 2000 pounds, the coal input can be calculated as \(100 \times 2000 = 200,000\) pounds per hour. - air is flowing into the combustion chamber at a rate of 1.3 million pounds per hour. 2. **outputs**: - ash is produced at a rate of 10 tons per hour, which equals \(10 \times 2000 = 20,000\) pounds per hour. - flue gases, which are the main subject of our calculation.

applying the conservation of mass, we set up the following equation: \[ \text{total input} = \text{total output} \] \[ \text{coal input} + \text{air input} = \text{flue gas output} + \text{ash output} \]

plugging in the known values: \[ 200,000 \, \text{lb/hr (coal)} + 1,300,000 \, \text{lb/hr (air)} = \text{flue gas output} + 20,000 \, \text{lb/hr (ash)} \]

to find the flue gas output, rearrange the equation: \[ \text{flue gas output} = 200,000 + 1,300,000 - 20,000 \] \[ \text{flue gas output} = 1,480,000 \, \text{lb/hr} \]

therefore, the mass rate of flue gases being exhausted from the plant is 1.48 million pounds per hour. this calculation ensures that the mass balance is maintained as required by the principle of conservation of mass, thereby confirming that the inputs and outputs within the system are equal.