100 mL of water at 15℃ is placed in the freezer compartment of a refrigerator with a coefficient of performance of 4.0. How much heat energy is exhausted into the room as the water is changed to ice at -15℃?

Video transcript

Hey, everyone. So this problem is dealing with cooling systems. Let's see what it's asking us. We have a cooling system with a coefficient of performance of 4.2 that's used to freeze a liquid weighing 2 kg initially at a temperature of 25 degrees Celsius. We're asked to determine the amount of heat energy exhausted into the room as the liquid solidifies at negative 10 degrees Celsius. The specific heat capacity of the liquid is 3.8 joules per gram. Kelvin and its heat of fusion is 140 Jews per gram are multiple choice answers. R A 2.66 times 10 to the five Jews B 6.76 times 10 to the five Jews C 2.80 times 10 to the five Jews or D 5. times 10 to the five Jews. So there are a few equations that we're going to need to recall to solve this problem. It's asking us for the amount of heat energy exhausted. So that's Q sub H which we can recall is equal to our work plus Q sub C which is our discharged heat. We aren't given work, but we are given the coefficient of performance. So we can recall that the coefficient of performance K is equal to of C or our discharged heat divided by our work. And so we can use a third equation to solve for Q sub C directly and then use these two to solve for Q sub H. So QC C we can recall is given by the equation MC delta T plus M multiplied by L sub F for the late heated fusion. And so this is our general heat transfer equation where we're looking at the heat transferred to bring our liquid down from at initial temperature of 25 degrees Celsius to this solid state at negative 10 degrees Celsius. And then because we have this phase change where we are solidifying from a liquid to a solid, we also have to include the phase change term, which is that M multiplied by RL sub f the laten heat of fusion. And so we are given all of these values in the problem. And so we can just kind of plug and chud from here. And it looks like the, when we're looking at the um values given, we have a number that are given in grams. And so we can choose to right our weight in grams as well. So that's what I or our mass. So that's what I'm going to do here. So 2 kg is the same as 2000 g. Our specific heat capacity, you see is 3.8 Jews per gram, Kelvin. And then our delta T is 35 degrees. So we're taking it from 25 C to negative 10 C. And when we look at our units, we do have degrees Kelvin on the bottom. But we can recall that a degree Celsius is the same as a degree Kelvin, the same unit size. And so a delta of 35 degrees Celsius is the same as a delta of 35 degrees Kelvin. So multiply, so those three terms are multiplied together, and then we're going to add this phase change term. So that's 2000 g again, multiplied by our laten heat of fusion where was given in the problem as 140 Joles per gram. So we plug that into our calculator and we get 5.46 times 10 to the five jos. And so now we have our Q sub C, we were told K in the problem. And so we can use our K um our co vision of performance equation to solve for our work. And so work is equal to Q sub C divided by K rearranging that equation to isolate work on the left side. And so Q sub C, we just saw 445.46 times 10 to the fifth Jews divided by K, just give it in the problem as 4.2. So that comes out to a work of 1. times 10 to the fifth Jules. And now finally, we can solve for our heat exhausted. So Q sub H is equal to our Q sub C 5.46 times 10 to the fifth Jews plus our work 1.3 times to the fifth Jules. And that comes out to 6.76 times 10 to the fifth. Sure. And so that is the final answer for this problem. And then when we look at our multiple choice answers, it aligns with answer choice B so B is the correct answer for this one. That's all for this problem. We'll see you in the next video.