An ice-cube tray of negligible mass contains 0.290 kg of water at...

Question

An ice-cube tray of negligible mass contains 0.290 kg of water at 18.0°C. How much heat must be removed to cool the water to 0.00°C and freeze it? Express your answer in joules, calories, and Btu

Accepted Answer

Hey everyone in this problem, we're squeezing oranges to make one cup of juice of mass 230 g at 25°C. Yeah. Were asked if the juice freezes at negative 10 degrees Celsius determine the heat that must be extracted from the juice to freeze it. Okay. And we're asked to find the value in jewels, calories and btu british thermal units. Okay. We're told to take the specific heat capacity in latent heat of fusion 3915 joules per kilogram Calvin and 30,016 kila jewels per kilogram respectively. Okay. All right, so let's get started. We're asked to find the heat. Okay, so we're thinking about heat, we're thinking about the value of Q. You know I find cute. And what is Q. Gonna consist of? Okay, well we're starting with juice at 25 degrees Celsius. Okay. And we want to freeze it at negative 10 degrees Celsius. Okay. So we first need to cool the juice. Okay? So we need a change in temperature and then we need it to freeze a negative 10 degrees Celsius. Okay? And that's going to be a phase change. And so we first need a change in temperature and then a phase change in order to go from juice at 25 degrees Celsius to freezing the juice at negative 10 degrees Celsius. Okay. And so Q. Is going to be equal to what? Well we have a change in temperature, recall that the equation is given by M. Okay. And in this case the mass. J. Of juice. See the specific heat capacity and delta T. Okay, so the mass of the juice, The specific heat capacity times times the change in temperature. Now the second term is going to be for the phase change. And when we have this phase change, the equation is going to be M. And again mJ the mass of the juice. And that times the latent heat. And in this case it's a latent heat of fusion. Because we're freezing and you'll notice that we have a negative sign here, we're subtracting. Now we're subtracting because we're losing heat in this phase change, our material is losing heat. Okay, when we're freezing the temperature is getting colder, the object or the material is losing heat. Okay, so we have that negative there. Alright, so let's go ahead. We're trying to find Q. And plug in the values that we know. Well, the mass of the juice we're told is 230 g. Okay, let's just work this out above. Okay, this is gonna be equal to 230 grams divided by 1000 g kilogram Which gives us 0.23 kg. Okay, so to go from g to kg, we divide by 1000. You got 0.23 kg. Okay, we're told see the specific capacity is 3, jules per kilogram Calvin. Okay, so we have 3915 jewels per kilogram Calvin. And then times delta T. R. Change in temperature. and because we have Calvin in our specific heat capacity we need our temperature DELTA T. To be in Calvin as well. And so delta T. Is going to be the final temperature which is negative 10 degrees Celsius plus 273.15 K. To convert to Calvin - the initial temperature 25°C. and again we add 273.15 to convert to Calvin. Alright and then I'm just gonna do the second term. This subtraction underneath because we're running out of room here we have the mass of the juice which again is 0.23 kg. And then we have the latent heat of fusion and we're told that that's 30,016 kila jewels per kilogram. Can we? So L. F. Is 30,016 k. J per kilogram. Okay we want to convert this into jewels per kilogram. And so what we're gonna do is we're gonna multiply by 1000. Okay we have kill a jewel. So to get to jewel we multiply by 1000. So this is going to be 30,016 times 10 to the three jules per kilogram. Okay times 1000. Alright so plugging that into our equation 30,016 times 10 to the three jewels per kilogram. Okay now if we work this out, the first term we have a negative 10 and the negative subtract 25 this is going to end up being negative. Okay, so that whole first term is going to be negative, we get negative 31,515. jewels. subtract this second term, which is going to be 600 And 90,000? No it's not, it's gonna be 6,903, jewels. Okay, And when we do this subtraction we get Q is equal to negative 6. four approximately times 10 to the six jewels. Alright, so the heat required Okay, to be extracted is going to be 6.94 times 10 to the six jewels. Okay, this negative indicates that it's being extracted And this value is in jewels. Okay, so this is part one and then for the other two parts we just need to convert this into calories in british thermal units. Okay, so for part two we want to convert to calories. Okay, so we have 6.94 and we're just gonna take the positive because we know that it's being extracted so we know we have that negative and so we're just looking for the magnitude of that value, so 6.9, 4 times 10 to the six jewels. Okay, and we have one calorie per 4.186 jewels which gives us a value of 1.66 times 10 to the six calories. And the final conversion into BTUs british thermal units, We have 6.94 times 10 to the six jewels times one Btu per 1055 jewels, which gives us 6.58 times to the three btu british thermal units. Okay, Alright. So if we go back up to our answer choices we see but we have answers. See okay, We have 6.94 times 10 to the six jewels, 1.66 times 10 to the six calories and 6.6 times 10 to the three BTUs. Thanks everyone for watching. See you in the next video.

An ice-cube tray of negligible mass contains 0.290 kg of water at 18.0°C. How much heat must be removed to cool the water to 0.00°C and freeze it? Express your answer in joules, calories, and Btu

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