heat lost by metal + heat gained by ice = 0
[mass metal x sp.h. x (Tfinal-Tinitial)] + [mass ice x heat fusiion)] = 0
25*X*(0-150) + (9g*333.5 J/g) = 0
Solve for X
A 25g piece of an unknown metal alloy at 150 degrees Celsius is dropped into an insulated container with 200 g of ice. Calculate the specific heat capacity of the metal, given that 9.0 g of ice melted.
How would you work this out?
2 answers
First you have to find the number of moles of water n=m/M = 9/18 = 0.05 mol
Latent heat of fusion of water is 6 kJ/mol, which means that 1 mole requires 6 kJ but you have only 0.05mol. Hence, the amount of energy required is 3kJ = 3000 J = q.
q(metal) = m*c*deltaT therefore c=q/mdeltaT=3000/(25*150)=0.80J/g0C
I hope it helps.
Cheers
JD (Chemistry tutor)
Latent heat of fusion of water is 6 kJ/mol, which means that 1 mole requires 6 kJ but you have only 0.05mol. Hence, the amount of energy required is 3kJ = 3000 J = q.
q(metal) = m*c*deltaT therefore c=q/mdeltaT=3000/(25*150)=0.80J/g0C
I hope it helps.
Cheers
JD (Chemistry tutor)
2 answers