Calorimetry 1

[tcarp77]

thanks for nothin, psycho bitch. trevor.

[007bankai]

thanks!

[cbarnes504]

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I LOVE U LADY I CANT SEE! IVE BEEN WORKING ON THIS PROBLEM FOR 3 DAYS NOW AND COULD NOT GET IT! THANK YOU SO MUCH THIS HELPED 100% :-) YAYYY!!!!

[ziggymrt]

how did you get to 23?....i tried and it didnt work

[Cscoped]

This video was very helpful, including the additional alegebraic calc post following. However, in the video after calculatin the final temp for the calorimeter, you then used this info to plant into the previous problem pertaining to Fe. So, my question has to do with the fact that there are 3 factors in this problem: qiron+qwater+qcal = 0. After performing the necessary calc, then am I to divide this value by 2 or 3 since the equation has 3 factors?

[chemistryprofessorpc]

I think your question refers to the last problem in that video. In that case, after you have multiplied and collected like terms, your final equation looks like this: 1299.2Tf = 29764

When you divide the 29764 by 1299.2, then Tf = 22.9deg C. You do not need to divide further by anything because all the factors are accounted for.

I hope this helps.

[addisballer]

Could you help me out this question.

-a small immersoin heater is tared at 350w,find the time t(min) to heat 250(ml) of water from 20 deg c to 50 deg c

thanks

[chemistryprofessorpc]

Wonder where my earlier response went??!! In any case, the item of information that you may be lacking is that 1 watt = 1 j/sec. So let's start out by finding how many joules of heat are needed. It takes 4.184j to heat 1 g of water 1 deg C and 50-20 = 30deg.

250mL x 1.00g/mL x 4.184 j/g-degC x 30degC = 31380 J of energy needed to heat up that quantity of water. Now, take that quantity and divide by 350j/sec to get the time. Watch your units cancel.

31380J x 1sec/350J = 89.7 sec or 90 sec.

[addisballer]

Thank you :)

[addisballer]

you are an amazing teacher. Thank you.

[bryceyap]

miss that should be -1265 Qcal right?

[chemistryprofessorpc]

Unless I have mess up, I get 1255.2 j, which then is divided by the 36 degrees C to arrive at 34.9 j/C - round off to 35 j/C as the heat capacity for the calorimeter. Do you find a math error? The net heat unaccounted for is about -1255 joules, therefore we say that is the heat absorbed by the calorimeter over a range of 36 degrees C. We want to find the heat absorbed per degree C, hence we divide. Is your concern that I may have mistyped the heat and put -1265 when I should have used -1255?

[chemistryprofessorpc]

Sure. Here's the set up. Remember, heat lost by the gold + heat gained by the water must equal 0.

MSdelta T(gold) + MSdelta T (water) = 0

3.0 kg x 0.129 j/gC x (Tf-99C) + 0.22 kg x 4.184 j/gC x Tf - 25C) = 0

0.387Tf - 38.313 + 0.9205Tf - 23.012 = 0

1.307Tf - 61.325 = 0

1.307Tf = 61.325

Tf = 46.9 degrees C

Remember, Tf is final temperature, Ti is initial temperature.  You might be less confused if you change 3.0 kg to 3000g and 0.22 kg to 220 g before you begin, so your units will cancel.

[lolahoneyforyou]

Miss chemistryprofessorpc, could you please help me solve this problem: "What is the final temperature when a 3.0 kg gold bar at 99 degrees celsius is dropped into 0.22 kg of water at 25 degrees celsius." your my last hope. Thank You :)

[chemistryprofessorpc]

The trick here is knowing how to deal with those negative numbers we have, sometimes. (20g)(0.450j/gC)(tf-440)+(300g­)(4.184j/gC)(tf-20)=0 Multiply out each section and get: 9tf-3960+1255.2tf-25104 = 0 Collect like terms: 1264.2tf - 29064 = 0 Now, add 29064 to both sides and get: 1264.2tf = 29064 Divide both sides by 1264.2 and get tf(final temperature) = 22.99 degrees C which I round off to 23 degrees C. Hope this helps!

[handbanana669]

Hello! My algebra isn't strong enough to solve the first Calormetry problem. Can you show me how I would get 23C???