a-how-much-energy-do-you-need-to-transfer-to-raise-the-temperature-of-150-mathrmg-of-aluminum-from-3

Question

a. How much energy do you need to transfer to raise the temperature of 150 $\mathrm{g}$ of aluminum from $20^{\circ} \mathrm{C}$ to its melting point?
b. How much energy do you need to transfer to melt 150 $\mathrm{g}$ of aluminum?

David Collins · Accepted Answer

Question eight takes into account something learned in this section and also the previous section. First, you&#x27;re asked to calculate the amount of heat that is required to raise the temperature of aluminum from 20 degrees Celsius to its melting point. This is using the traditional heat equation that was provided in the previous section, where heat is equal to the specific keep capacity of aluminum multiplied by its mass multiplied by its change in temperature. So you&#x27;ll need to go back to the previous section and look up the specific heat of aluminum. It&#x27;s point to one calories per gram degree. Celsius will multiply that by the mass 150 g and then multiply that by the change in temperature. The final temperature, which is the melting point of aluminum, which you will look up in the table in this section 659 degrees Celsius, minus the initial temperature 20 degrees Celsius. That then gives us ah heat of 20,100 calories. Now that it it is at its melting point. How much heat is required to melt? 150 g? This then refers to what you learned in this section, where you simply use the heat of fusion to convert the grams of aluminum into the calories of aluminum required to melt it at its melting point, which is three. Heat of fusion of aluminum is 94.5 calories per gram. This then gives us a value of 14,200 calories.

David Zhang · Answer

So for this question, we have Q because use this equation. For the first part, we have passed 3000 grams, which is three kilograms. We have especially heat up to a 0.9 or aluminum and changing temperature is 660 minus 20. Get the math for this issue. Queue them out of, um, energy you need is 1.73 times 10 to the sixth Stools party. Never part B one used something similar equation use the same mess actually can use. We can just use the kilogram version because the letting heat fell. You&#x27;re gonna use uses. Killer jewels were for kilograms. So the mass has three kilograms because he is 397 killer jewels per kilogram and you should get that amount of energy you need is around 1190 killing jewels.

Dading Chen · Answer

So the question was asking us finding energy. They need to raise the tempter of Tokyo Grand chunk of a little number 18 these charges. So we know. Ah, the equation for such can question is qu goto emceed hostility. Here&#x27;s the heat. Oh, in this case, the energy that acquire M is the mass sees us of the city heats and delta T is the change in temperature. So, uh, since we know all these conditions for a mass simplicity and the change in temperature of Uganda simplicity off the alumina is Nash in 21 point. Oh, nice six. You per kilograms times calving. Okay, so cute can be just simply equal to the mass which is came as 2.4 kilograms times 121.96 ju pro kilograms times Calvin times The changing temperature, which is 18 researches and this will give us the heat&#x27;s on the energy they need to raise Tokyo grand chunk of Louima. Remember eating the results? It is about three point three times potential of power afford ju Okay, if you want the writing in killed you is 33 Children&#x27;s can&#x27;t and this is the answers for the questions. Thank you

Kevin Zaborsky · Answer

we want to determine how much heat energy Q is required to heat up 55 grams of aluminum from 22.4 degrees Celsius to 94.6 degrees Celsius. We can use the equation. Q equals M C Delta T in order to solve for Q, since we&#x27;re given values free Other variables in this equation so cue equals M. So the massive aluminum, which is given as 55 grams R M cp, which is a specific heat capacity of aluminum, and we can look up that value to be 0.8 97 Jules, divided by grams times Kelvin. And then we multiply that by the change in temperature Delta T, which will always be the final temperature minus the initial temperature. So that&#x27;s 94 0.6 degrees Celsius minus 22.4 degrees Celsius. We went to cancel off units of Kelvin from the specific heat capacity, but we do not need to convert the temperature units from degrees Celsius to Kelvin because we&#x27;re finding a difference in temperature. So if we were to add 273 to both of the temperature values and subtract the initial from the final temperature, we would still get the same difference. And that&#x27;s why these units would cancel out grams cancel out to yield units of energy for Q. When we multiply those three quantities together, we should get a final answer of Q to be 3.6 times 10 to the third Jules of heat energy, required to you increase the temperature of that amount of aluminum.

Farouq Hanonn · Answer

Okay, back again. Let&#x27;s do this question. And for part A, we want to find the heat transfer required to raise the temperature of aluminum part that aluminum part. Let&#x27;s say this is our part here. And it has a mass of 0.750 kilograms and it contains 2.5 old. So for water, you blue. The water is gonna have a mass of specifically 2.5 kilograms. So this part has water and the pot is made out of aluminum. And we want to know that he transferred, required to raise the temperature from third degree Celsius to the boiling point. Then we boil away 0.75 old kilograms of water. So what&#x27;s going on here is we are the first stage is we take 30 degrees Celsius and we go to the boiling point. We boil ving we boy away step number two working boy aways your 20.75 kilograms of water and this is going to be boiled. So when you know that we will have some change face change in this example. So we&#x27;re gonna have Q equals m times l and we know we have the latest he a fusion, which is our else of F and the late lady Vaporization, which is our l sub d. So these are important equations and we know the heat transfers, Q equals and so we&#x27;re going to use these equations at her part. A. We&#x27;re going to see Q equals that m cat, which is gonna be m c Delta T for water. So we take the mass for water, see for water Delta T. So that&#x27;s her first term. Then we&#x27;re gonna add that mass for aluminum, See, for aluminum and delta t for that different change in temperature. So we have both water and aluminum. They&#x27;re going to give up. He we&#x27;re gonna have some form of heat transfer. Then the last thing we have to add is the massive water times l. So it&#x27;s plus the massive water times l is the else of F or else a V. Do we have fusion or vaporization here? It&#x27;s going to be vaporization because we&#x27;re looking at boiling. It&#x27;s gonna get vaporised. So right gonna have V. So we know we&#x27;ve accounted for the face change at the end for the Q equals M l. And we&#x27;ve accounted for the water he transfer and aluminum heat transfer. So now we can solve for accuse. So the massive water is going to be 2 25 old kilograms the specific heat for water and the specific heat for water. We know we are final answer to be in K cow. So when you wanna use 1.0, que cow her kilogram degrees Celsius. So we have that we need the calories in this. So it&#x27;s killer calories because we are final answer. They&#x27;re quite he transferred in, kill CalPERS Then we have the change in temperature change in temperature. We take the boiling point of water which is 100 degrees Celsius, right and we start at 30 degrees Celsius. So we subtract there. That&#x27;s our change in temperature this time. Then we do plus Now we&#x27;re looking at the massive aluminum. The massive aluminum is going to be 0.7 bike. Oh, kilograms. That&#x27;s right. Then we have the average specific key for aluminum. This is found in your book in 0.215 kill Cal her kilogram degree Celsius. Then we take our 100 degrees Celsius minus Terry. So that does not change. So it&#x27;s 100 degrees Celsius minus 30 which is our delta t. So that&#x27;s that. So we were done with this term. Now we are at the last term. So we do. Plus 10 5 old. That&#x27;s the water. We want a boil away. Don&#x27;t get confused with the original Massive the water in the question it 0.5 kilograms of water. We want a boy away. Then we multiply. Borrow Elsa. Be so this is going to be the lady He vipir ization vaporization for the water. It&#x27;s gonna be 5 39 Kill a cow Her kilogram. This is found in table in our book. So now we can do the mass For all of this you can solve for accuse You should get around 5 90 54 killed, Cal So this is our answer for part A For part B. What we want to find essentially is going to be how long it&#x27;s gonna take for the rate. If you transfer is when it&#x27;s 500 watts, how long will this take? So what we&#x27;re gonna do, we know Q equals P Times T, and we can solve for the time and the power is gonna be 500. Well, so we just plug in 500 watts for the power, and we found our Q value, right? It&#x27;s the 5 90.54 So t equals Q over P and our Q values going to be the 5 90 54 Kelkal. And we know we need to convert this to Jules, so we know that there is going to be four 186 jewels and one kill. Okay, so we have to convert that cube. Then we divide all of this by the power, which is 500 walks. So at the end, we know we can solve for the time here, and you should get around 4.94 times 10 to the three seconds. So this is our answer for part Big? I don&#x27;t know.

Sheh Lit Chang · Answer

this question. We have a liminal part. We insight. Yeah. Think so. We want to find a ah um part of one. Well, you know. Yeah. Until 1.1 one. Right. Right. I don&#x27;t quite so identifier about heating. No. Ah Right. Job We see 200 BC. Thank you. Don&#x27;t want to higher degree C. So it&#x27;s pointing away. You also want to find out how much time his own and. Mhm. Okay. So we need you. Uh huh. Uh huh. Yeah. Mm hmm. Mhm. The you know and what? All right. The most you know. Cool. 800 you. Mhm. Crazy. Uh huh. The heat capacity of what? Yes. For what? It&#x27;s sick juice per kg. Mhm. Then heat of vaporization of Yeah. To fight you know jews. We need to use these numbers. Mhm mm. That he required. Mhm. So uh we are going to use your form but you request to and see you know for hitting mm mm. Then the other formula is Q. It goes and lg well coiling mm. So uh the the energy required. You Yeah. You got water last accu without the aluminum parts. He plants the queue for boiling off. What the hell? So yeah. Uh huh. Q. For heating up water. Mhm mm. And W. C. W. Yeah. Mm. Yeah. Data. That&#x27;s our and mm. See aluminum you know. Uh huh. Mw. Of the So just putting the numbers five times 1 is sick Times 70° structures. The Math of aluminum part 0.5 Kg. 900 mm. Ah Right. You want right on. And I multiplied by. Sure. Thanks understanding the tree. Mm. And you&#x27;re all this up 47 Okay. Thanks. Yeah. Yeah. So this now he needed. Mhm. Phone it&#x27;s been described in party then in part B. I want to find the time taken thinks will be using formula. Yeah. To our time. Mhm. Yeah. Mhm. You divide E K. Your P. And so the Q is 24 7 times, right? How is 500? And get 4940? We went minute. You get mhm. Cool, mm. Mhm. Yeah. Yeah. And that&#x27;s all.

Problem

You place 25.0 $\mathrm{g}$ of ice at $0^{\circ} …

Problem 8 Easy Difficulty

a. How much energy do you need to transfer to raise the temperature of 150 $\mathrm{g}$ of aluminum from $20^{\circ} \mathrm{C}$ to its melting point?
b. How much energy do you need to transfer to melt 150 $\mathrm{g}$ of aluminum?

Answer

a) 20128.5 $\mathrm{cal}$
b) 14175 $\mathrm{cal}$

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a) 20128.5 $\mathrm{cal}$b) 14175 $\mathrm{cal}$

Living by Chemistry

Stephanie C.

Lizabeth T.

Karli S.

Jake R.

Liquids - Intro

Comparison Of Gases Liquids And Solids

Angelica M. StacyLiving by Chemistry

Stephanie C.

Lizabeth T.

Karli S.

Jake R.

a) 20128.5 $\mathrm{cal}$
b) 14175 $\mathrm{cal}$

Angelica M. Stacy

Living by Chemistry