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Calculate the amount of heat absorbed when 5.50 g of aluminum is heated from 25.0ºC to 95.0ºC. The specific heat of aluminum is 0.897 J/(g?ºC)...

Question

Calculate the amount of heat absorbed when 5.50 g of aluminum is heated from 25.0ºC to 95.0ºC. The specific heat of aluminum is 0.897 J/(g?ºC)

Calculate the amount of heat absorbed when 5.50 g of aluminum is heated from 25.0ºC to 95.0ºC. The specific heat of aluminum is 0.897 J/(g?ºC)



Answers

A 25.5-g aluminum block is warmed to 65.4 C and plunged into an insulated beaker containing 55.2 g water initially at 22.2 C. The aluminum and the water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum?

So here we just have 25.5 kg of a block, but it's warm to 65.4, placed into a beaker containing 55.2 g of water that is initially 22.2 degrees Celsius. We reach Samel equilibrium, and we're just determining the final temperature of both. So is he transfers from one substance to another. Que must be equal in magnitude, but opposite in direction freaks substance tense. Why we have the exact same temperatures present. So here we will expand both sides of the equation so that we include variables of known values that are listed above in the description and then following this, you just simply sold for your unknown temperature. So this is the equation that I expand out, and then I incorporate my variables, which generates a value off 26.1 degrees Celsius.

Question 36 has supposed 72 point three. Jewels of heat are added thio 101 gram piece of aluminum. So that tells us our specific heat. If we go look in the table. The specific heat of aluminum is 900 jewels per kilogram. Kelvin on. We'll come back to that unit in a minute. Jules per kilogram. Calvin on it says the piece of aluminum is initially at 20.5 degrees Celsius. Were supposed to find the final temperature of the aluminum. Remember, Q E equals M C. Delta T, but we do need to make sure our units match. That's in jewels. That's Ingram's. That's in jewels per kilogram, Kelvin. So we either need to change this to grant kilograms or this two grams that conversions the same. We move our decimal 0.3 places. We'll just go ahead and say this is 0.101 01 kilograms. And so now I consult for Delta T. So 72.3 is equal to M 100 or 1000.101 one a one time see 900 Delta T. And so will divide 72.3 by 0.101 and nine, hundreds of divide by 90.101 times 900. Divide by point. One on one and 900 0.1 a. One times 900 we see 9000.795 So our delta T is 0.795 degrees Celsius. Now I know that this unit says Per Kelvin, remember, that's a change of a Kelvin. And a change of one Calvin and a change of one degree Celsius is the same. So if this started at 20.5 and the increased by 0.795 plus 0.795 then all we have to do is add those together and we see our final temperature is 21.295 degrees Celsius, okay?

Hello Today we'll be talking about Chapter 15 Question 10 which asks us to consider a piece of aluminum that that's heated to about 95 degrees Celsius. So we have a 5.50 graham block of aluminum that goes from 25 degrees Celsius all the way to 95 degrees Celsius. That's a 70 AA degree temperature change. So that's our delta T. And were asked to determine how much heat was needed to cause this change. And so if we write out our equation for heat and temperature change, we have. Q. The heat absorbed. Curious heat equals the mass of the block. There's the mass, this specific heat and lastly, the temperature change. And we can look up the specific heat of aluminum in a table, and we find that the specific heat is 0.897 point 897 Jules, her cram degree Celsius of aluminum. And so with that, we're all set to solve for Q and start plugging our numbers into the equation, so Q equals 5.50 grams aluminum times 0.897 jewels per gram of aluminum, her degree Celsius degrees, Celsius times 70 degrees Celsius. We can see that our units are grams of aluminum. Cancel as to our degrees Celsius. That leaves us with jewels as our final answer. And indeed, we find that we have 320 0.7 jewels of energy that had to be added to this aluminum to raise the temperature. And so if we correct for significant figures, we have to round this to 3.21 times 10 to the second Jules because we have two significant figures in our mass or three significant figures in our mass, but for in the number that we spat out. And so hopefully this helps you become a little bit more comfortable with using the equation to relate, heat absorbed and temperature change.

So in thermal equilibrium in the final temperature off aluminum block and water should be the same. So let's assume the final temperature is t so that the heat gives off by aluminum. Block equals to emcee a 65 65.4 minors T in which the heat capacity is the subsidy. He'll capacity of aluminum on the cure. Watery equals two and C T minus 22.2. This is Dr T. On this sea is the subservient heat capacity of water. So the cure water he wants to cure aluminum block. So based on this two equations T host to 26.1 Celsius degree.


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