what is the mass of a lead block that if heated to 100.0 dgrs C and placed in a container with 50.0g of ice at 0.00dgrs C produces a final water sample of 15.0dgrs C? Sh of Pb= 0.128J/g dgrsC step by step solution and answer
q1 for Pb block = mass x sp.h. x (Tf - Ti).
mass Pb = X
sp.h. Pb = listed
Tf = 15.0
Ti = 100.0
q2 for ice at zero degrees to melt to water at zero degrees is mass x delta H fusion. Be sure to use J/g for the unit.
mass = 50.0 g
delta H = ?? J/g.
q3 for water at zero to be heated to 15.0 degrees = mass x sp.h. x (Tf - Ti)
mass H2O = 50 g.
sp.h. water = ?? J/g*C.
Tf = 15.0
Ti = 0.0
q1+q2+q3 = 0
solve for X.
The sum of the heats gained is zero (some lose heat).
Heat gained lead+ heatgainedice=0
m*Cpb*(100-Tf)+ 50*Heatfusionice + 50*cwater*(Tf-0)=0
you are given Tf, solve for m.
2 answers