Description
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?
Explanation & Answer
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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
538.2=15.4808m
m=538.2/15.4808 =34.77 gram.(answer)
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