Help with chemistry problem

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A 32.5 g Iron rod, initially at 22.7 degree C, is submerged into an unknown mass of water at  63.2 degree C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.5 degree C. What is the mass of the water?

Nov 3rd, 2015

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We have

Mass of iron=32.5 gram

Specific heat of iron=.45J/g degree C

Change in temperature=59.5.2-22.7=36.8 degree C

Heat gained =mCdT =32.5*.45*36.8 J=538.2 J

Now for the water

Mass to calculate

Specific heat of water =4.184 J/g degree C

Change in temperature=63.2-59.5=3.7

Heat released by water=mCdT=m*3.7*4.184 J=15.4808 J

As the heat gained by the iron is at the expense of heat released by water therefore

Equate both and solve for mass of water


m=538.2/15.4808 =34.77 gram.(answer)

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Nov 3rd, 2015

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