Chemistry Problem 2?

Question

A 500 g Iron bar at 212 degrees celsius is placed in 2 liters of water at 24.0 degrees celsius. What will be the change in temparature of the water?(Assume no heat is lost to the surrounding) ( Specific heat of Fe = 0.493 J/g Degrees Celsius)

Answer 1

If no heat is lost to the surroundings, then all of the heat is absorbed by water. Therefore:
Qw = QFe
mwCpw(T2w-T1w) = mFeCpFe(T2Fe-T1Fe)

and T2w = T2Fe = T2 (at the end they're at equilibrium)

(2 l *1000 g/l)(4.184 J / g)(T2-24)=(500g)(0.493 J/g)(T2-212)

Just solve for T2

Answer 2

suppose the final temperature is Tf

You can say the absolute amount of heat lost by iron will increase the temperature of water

amount of heat lost by iron
Wi= -(Tf -212)*500*0.493
amount of heat received by water

Wh = (Tf-24)*2000*4.18
since 2l of water =2000g and 4.18 J/g specific heat of water

so absolute value Wh=-Wi

(212-Tf)*500*0.493= (Tf-24)*2000*4.18
(212-Tf)*0.493 = 16.72(Tf-24)
104.516-0.493Tf= 16.72Tf -401.28
505.796= 17.213Tf
Tf=29.4°C