A thermally-jnsulated container made from 120 g of aluminum contains 124 g of wa

Question

A thermally-jnsulated container made from 120 g of aluminum contains 124 g of water at 20degreeC. 40.0 g of ice at 0degreeC are added to the water in the container. The contents are then gently stirred to keep the temperature of the system uniform. Lt = 333 J/g, cw = 4.186 J/(g . K), cAI = 0.900 J/(g . K) The water equivalent, meq of the container's the mass of water that would have the same heat capacity Q/DeltaT as the container. Show that the water equivalent of the container is meq = 25.8 g. (Q = mcDeltaT.) Does all the ice that was placed in the container melt? (The effective mass of water in the container is mw= 124 g + meq.) (You must show an appropriate calculation.) Determine the final temperature of the system. If all the ice does not melt, determine how much ice remains after the system has achieved thermal equilibrium. (Show work; you will not be given credit unless you show how you calculated your answer.)

Explanation / Answer

if meq is mass of water equivalent

then meq= mal x cal / cw= 120 x 0.9 / 4.186= 25.80 g

b. mw= 124 + 25.8= 149.8 g

heat required for 40g ice to melt = 40 x 333=13320 J

heat rejected by water to reach 0 C = 149.8 x 4.186 x 20= 12541.25 J which is less

hence all ice will not melt

c. final temp will be zero

amount of ice melt= 12541.25/333= 37.66 g

amount of ice remaining = 40 - 37.66 = 2.34 g

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