#10 on Hotdog

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dgerges 4H
Posts: 65
Joined: Fri Sep 28, 2018 12:24 am

#10 on Hotdog

Postby dgerges 4H » Tue Feb 12, 2019 2:58 pm

10. Michael asks me for water with no ice at a dining hall. Just to spite him, I measure 25.0 g of ice at 0.00 oC and drop it into 265 mL of water at 25.0 oC. What is the final temperature of the water?
i know we set qice=-qwater and from that i got a T of 296.91K. Now to find the final answer do I calculate the final T using the enthalpy of fusion of water when the ice melted? This gives me 290K and the answer is supposed to be 289.
thanks for any help!

Jennifer Su 2L
Posts: 47
Joined: Wed Nov 21, 2018 12:20 am

Re: #10 on Hotdog

Postby Jennifer Su 2L » Tue Feb 12, 2019 4:33 pm

q(ice) = -q(water)

FOR ICE:
Melting: q=deltaH(fusion)= 6.01 kJ/mol x (25.0 g / 18 g*mol-1) = 8347.2 J
Increasing Temp of Melted Ice: q=mCdeltaT= (25.0 g) x (4.184 J/K*g) x (Tf-273K) = 104.6(J/K)*Tf - 28555.8J

FOR WATER:
Cooling the Water: q= mCdeltaT= (265 g) x (4.18 J/K*g) x (Tf-298K) = 1107.7(J/K)*Tf -330094.6J

Since q(ice)=-q(water)
[(8347.2 J) + (104.6(J/K)*Tf - 28555.8 J)] = -[1107.7(J/K)*Tf -330094.6 J]
1212.3(J/K)*Tf = 350303.2 J
Tf= 288.957K which rounds to three sigfigs: 289K

dgerges 4H
Posts: 65
Joined: Fri Sep 28, 2018 12:24 am

Re: #10 on Hotdog

Postby dgerges 4H » Tue Feb 19, 2019 10:41 am

thank you!!

avshi10b
Posts: 6
Joined: Mon Apr 30, 2018 3:00 am

Re: #10 on Hotdog

Postby avshi10b » Sun Mar 17, 2019 7:46 pm

Thank you


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