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Chapter 10: Problem 125

In regions with dry climates, evaporative coolers are used to cool air. A typical electric air conditioner is rated at \(1.00 \times 10^{4} \mathrm{Btu} / \mathrm{h}\) (1 Btu, or British thermal unit \(=\) amount of energy to raise the temperature of \(1 \mathrm{lb}\) water by \(\left.1^{\circ} \mathrm{F}\right) .\) What quantity of water must be evaporated each hour to dissipate as much heat as a typical electric air conditioner?

Short Answer

Expert verified
The quantity of water that must be evaporated each hour to dissipate as much heat as a typical electric air conditioner is approximately 10.31 lb/h.

Step by step solution

01

Calculate the rate of energy absorption in Btu/h by the evaporative cooler

To dissipate heat at the same rate as the electric air conditioner, the evaporative cooler must absorb energy at a rate of \(1.00 \times 10^{4}\) Btu/h.
02

Determine the energy requirement to evaporate 1 lb of water

As given, 1 Btu is the energy needed to increase the temperature of 1 lb of water by \(1^{\circ}\)F. To convert to the energy required to evaporate 1 lb of water, we know that the latent heat of vaporization of water is approximately 970 Btu/lb. Therefore, 970 Btu of energy is needed to evaporate 1 lb of water.
03

Calculate the quantity of water to be evaporated

Now, let's denote the mass of water to be evaporated per hour as \(m\), measured in pounds (lb). We'll use the relationship between energy, mass, and latent heat of vaporization to find the mass: Energy = mass × latent heat of vaporization \(1.00 \times 10^{4}\) Btu/h = \(m\) × 970 Btu/lb
04

Solve for the mass of water

To solve for the mass of water, we can simply divide the energy absorbed by the latent heat of vaporization: \(m = \frac{1.00 \times 10^{4}\;\text{Btu/h}}{970\; \text{Btu/lb}}\) \(m \approx 10.31\) lb/h
05

Final Answer

The quantity of water that must be evaporated each hour to dissipate as much heat as a typical electric air conditioner is approximately 10.31 lb/h.

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Most popular questions from this chapter

Mn crystallizes in the same type of cubic unit cell as Cu. Assuming that the radius of \(\mathrm{Mn}\) is \(5.6 \%\) larger than the radius of \(\mathrm{Cu}\) and the density of copper is \(8.96 \mathrm{~g} / \mathrm{cm}^{3}\), calculate the density of \(\mathrm{Mn}\).

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