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#10 on Hotdog

Posted: Tue Feb 12, 2019 2:58 pm
by dgerges 4H
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!

Re: #10 on Hotdog

Posted: Tue Feb 12, 2019 4:33 pm
by Jennifer Su 2L
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

Re: #10 on Hotdog

Posted: Tue Feb 19, 2019 10:41 am
by dgerges 4H
thank you!!

Re: #10 on Hotdog

Posted: Sun Mar 17, 2019 7:46 pm
by avshi10b
Thank you