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if 2500 ft3 of air is crossing an evaporator coil and is cooled from 90 f/32 c to 70f/21c. what would be the volume of air, in ft3 (cubic feet), exiting the evaporator? (whose equation? charles ? v1/t1

User Jmdecombe
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Final answer:

Using Charles's Law, we can calculate the volume of air exiting an evaporator after being cooled from 90°F to 70°F. By converting temperatures to Kelvin and applying the law's formula, we find that the volume decreases as the air cools.

Step-by-step explanation:

The question deals with the application of Charles's Law of gases, which states that at constant pressure, the volume of a gas is directly proportional to its absolute temperature (measured in Kelvin). To find the volume of air exiting the evaporator after cooling from 90°F/32°C to 70°F/21°C, we will use this principle.

Charles's Law is described by the formula V1/T1 = V2/T2. First, we convert the Celsius temperatures to Kelvin by adding 273.15 to each. So our temperatures become: T1 = 32°C + 273.15 = 305.15 K and T2 = 21°C + 273.15 = 294.15 K. Using the initial volume of V1 = 2500 ft3 and T1, we solve for the unknown V2 at the new temperature T2.

The equation becomes (2500 ft3 / 305.15 K) = V2 / 294.15 K. Solving for V2, we find the new volume of air exiting the evaporator coil. Because the temperature decreases, we expect the volume will also decrease according to Charles's Law.

User TaoPR
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