A 208 -g aluminum calorimeter contains 648 g of water at 18 ° C. A 105 -g piece

Question

A 208-g aluminum calorimeter contains 648 g of water at 18° C. A 105-g piece of ice cooled to -18° C is placed in the calorimeter. (Assume that the specific heat of ice is always 2.02 kJ/kg · K.) (a) Find the final temperature of the system, assuming no heat is transferred to or from the system.
1° C

(b) A 180-g piece of ice at -18° C is added. How much ice remains in the system after the system reaches equilibrium?
2 g
(c) Would the answer for Part (b) change if both pieces of ice were added at the same time?

Explanation / Answer

a) heat needed for 105g of ice melted completely. Q=105*18*2.02+105*333=3.88e4(J) amount of heat given out by aluminum and water at 18C when its temp is zero Q'=208*0.9*18+648*18*4.2=5.24e4(J). Q'>Q so the ice will be melted completely. let T be final temp. Q+105*T*4.2=(208*0.9+648*4.2)*(18-T) so 3350T=1.36e4 so T=4(deg). b) using the same method. Q=180*18*2.02+180*333=6.65e4(J). now Q>Q'. so the final temperature is zero. amount of ice become water. m*333=Q'-180*18*2.02 so m=138(g). c). yes.

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