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Equilibrium temperature of ice, water, and iron cup 
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#1
Nov2606, 11:18 PM

P: 99

so there're:
0.041 kg ice cube at 0.0 *C 0.110 kg water at 40.0 *C in 0.062 kg iron cup at 40.0 *C Find the equilibrium temperature of the cup and its contents:  So what I tried doing was... I found the heat lost by water if cooled to 0 *C, which was 18418.4 J (mass of water * water specific heat * temperature difference which is 40) then found th eheat needed to melt ice, which was 13735 J (mass of ice * Lf) then found the difference of the two to find the amount of heat left... and used that much heat to warm 0.151 kg of water (mass of initial water + initial ice) at 0 *C to its final temperature... so I did, delta T=Q/(mc) which gave T=7.4 *C... so then I used that temperature in kelvin, which is 280.6 K and did: (specific heat of water)(mass of water which is now 0.151)(280.6+T)=(specific heat of iron)(mass of cup)(313T) and solved for T.... where did I make the mistake because this answer's not correct? 


#2
Nov2706, 08:47 AM

P: 4

i think you can assume the equilibrium temperature is T , the three objects will reach this temperature after they are in eaulibrium
and the heat will be released from cup and water which in turn melt the ice so Q(ice get) = Q(cup release)+ Q(water relesase) Q= CM(Tt) Plut every value in the equation you will find out the final T try and see whether it is right 


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