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A 32.5 g cube of aluminum initially at 45.8 oC is submerged into 105.3 g of...

A 32.5 g cube of aluminum initially at 45.8 oC is submerged into 105.3 g of water at 15.4 oC. What is the final temperature of both substances at thermal equilibrium? The specific heat of Al is 0.903 J/g • oC. The specific heat of water is 4.184 J/g • oC. (Hint: your answer should not contain any units and have three significant figures)

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Answer #1

solution:

Heat lost by aluminium = heat gained by water

Mass of aluminium* specific heat of aluminium * (45.8-T)= mass of water* specific heat of water*(T-15.4)

specific heat of aluminium =0.91 j/g.K specific heat of water= 4.18 j/g.k

32.5*0.91*(45.8-T)= 105.3*4.18*(T-15.4)

T= (32.5*0.903*45.8+105.3*4.184*15.4)/ ( 32.5*0.903+105.3*4.184) =17.29 deg.c


answered by: ANURANJAN SARSAM
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