Question
In: Chemistry
2. A 155 g sample of iron metal at 120.0C is placed into 250.0 g of...
2. A 155 g sample of iron metal at 120.0C is placed into 250.0 g of water at 20.0 °C in a calorimeter. When the system reaches thermal equilibrium, the temperature of the water in the calorimeter is 30.8C. Assume the calorimeter is perfectly insulated. What is the specific heat of iron? What is the amount of heat for 80.0 g piece of iron?
Solutions
Expert Solution
Heat lose of Iron = heat gain of water
mc
t
= mct
155*C*(120-30.8) = 250*4.18*(30.8-20)
155*C*89.2 = 250*4.18*10.8
13826*C = 11286
C = 11286/13826 = 0.816j-C0/g
What is the amount of heat for 80.0 g piece of iron
q = mCt
= 80*0.816*(120-30.8)
= 80*0.816*89.2
= 5822.976joule
queen_honey_blossom answered 6 months agoRelated Solutions
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