Home air conditioners in the United States have their power specified in the truly obscure units of tons, where 1 ton is the power needed to melt 1 ton (2000 lb or 910 kg) of ice in 24 hours. A modest-size house typically has a 4.0 ton air conditioner. If a 4.0 ton air conditioner has a coefficient of performance of 2.5, a typical value, at what rate in kW is heat energy removed from the house?

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Accepted Answer

Hey, everyone. So this problem is dealing with cooling cycles. Let's see, what is that for us? We're told to consider that one ton is the amount of heat required to melt one tonn of ice in 24 hours. And they define one ton as £2000 or kg. A small scale storage facility has a 2.5 ton refrigeration system with a coefficient of performance of 2.7. Assuming the refrigeration system operates at its typical performance. What is the rate of heat energy removal from the storage in kilowatts? Our multiple choice answers are a 3.3 kilowatts, B 5.2 kilowatts C, 11 kilowatts or D 7.1 kilowatts. And so the key to solving this problem is recognizing that the rate of heat energy removal is the power. And so we can recall that power is equal to work divided by delta T or a change in time. Our work in turn can be found using the equation for coefficient of performance which we use to, we use K to denote and so K is equal to Q sub C which is our discharged heat divided by W our work. And then lastly, we have Q sub C for ice where we're, where all we're doing is a phase change. And so our Q sub C is equal to M multiplied by L sub F or our latent heat of fusion. And so if we rearrange this equation to uh if we arrange our co vision of performance equation to isolate work on the left hand side, we have work is equal to our Q sub C, our discharged heat divided by K and then Q of C is equal to our mass multiplied by our late heated fusion. And so work is equal to M multiplied by alsa be divided by K. And then we can plug that into our power equation. So we'll have powers equal to M multiplied by alpha F divided by K multiplied by delta T. And from there, we have a plug and chug because we have all of our values for the problem or known constants. So the mass we have 2.5 tons and they actually give us that conversion factor in the problem. And so that is 910 kg per ton. We wanna keep everything in standard units. And then the latent heat of fusion of ice is a constant. We can recall 3.34 times 10 to the fifth Jews per kilogram divided by our coefficient performance. Given in the problem is 2.7 multiplied by our delta t so change in time. So we're talking about a 24 hour time span. So 24 hours and then keeping standard units, we have 3600 seconds per hour as our conversion factor. So we plug that into our calculator and we get 3.3 times 10 to the three watts, which is equal to 3.3 kilowatts. So that is the correct answer for this problem. And it aligns with answer choice. A that's all we have for this one. We'll see you in the next video.

Verified Solution

Key Concepts

Coefficient of Performance (COP)

Tons of Refrigeration

Heat Transfer