8.41

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camrynpatterson3C
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8.41

Postby camrynpatterson3C » Sun Jan 14, 2018 5:06 pm

What formula are we supposed to use to find the heat changes of the ice cube and the water? And why do we set the equations equal to each other?

Joshua Hughes 1L
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Re: 8.41

Postby Joshua Hughes 1L » Sun Jan 14, 2018 5:22 pm

The equations for the heat of the water of the ice and the heat of the surrounding water should be set equal to each other because the two temperatures will approach an equilibrium (which should be 31 degrees C), ice warming up and melting and the surrounding water will cool down because of the ice.

Michaela Capps 1l
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Joined: Thu Jul 27, 2017 3:00 am

Re: 8.41

Postby Michaela Capps 1l » Sun Jan 14, 2018 5:26 pm

You have to solve by melting the ice, and then by bringing the liquid to the final temperature (which is endothermic/requiring heat). Heat absorbed by ice cube = (50g/18.02g*mol )(6.01*10^3 J*mol^-1) + 50g*4.184J*C^-1 *g^-1 * (Final Temperature-0 degrees C)
Heat released by liquid water = (400g)(4.184J*C^-1*g^-1)(Final Temperature - 45 degrees C)
Heat absorbed + Heat released = 0
Solve for final temperature : T = (5.8 x 10^4 J)/(1.88 x 10^3 J*C^-1) = 31 degrees C

Ricardo Ruiz Flores 1D
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Joined: Sat Jul 22, 2017 3:00 am

Re: 8.41

Postby Ricardo Ruiz Flores 1D » Sun Jan 14, 2018 5:29 pm

Our goal is to find final temperature, so we can use the q= ΔH=nCΔT because ΔT includes initial and final temperatures inherently.

The solution manual sets the ice cube and water's heats equal because it's explained that the heat gained by the ice cube will be equal to the heat lost by the hot water. Once we find both heat values, and because final temperature will be the same, we can leave final temperature as a variable, set them equal, and solve for it.

Joshua Hughes 1L
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Re: 8.41

Postby Joshua Hughes 1L » Sun Jan 14, 2018 5:35 pm

The equations be used I think is Q = mcΔT
q (heat) =(mass in grams)(Specific heat capacity)( Tfinal-Tinitial) They had to convert the mass from grams to moles because the specific heat they used wasn't in terms of grams (they got the specific heat from one of the tables in the chapter)
The reason the work shown in the solutions manual for solving for Q of the ice is that there are two parts to the temperature change of the ice. The heat needed to melt ice at 0 degrees C and then the heat needed to raise the temperature from 0 to its final temp. The surrounding water just cools down and doesn't have any change in state. We set them equal to each other as I explained earlier and then go about solving for Tfinal. Hope this helps and just say if you need additional clarification, I'm not super good at explaining this sorry


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