Question #f09ab

We know that in such interactions heat is gained by one and heat is lost is by the other.

Also that the heat gained/lost is given by

#DeltaQ=mst#, or #DeltaQ=mL#
where #m,s and t# are the mass, specific heat and rise or gain in temperature of the object;

#L# is the latent heat for the change of state and

#Delta Q_"lost"=Delta Q_"gained"#

In the given problem heat is lost by steam and gained by water aluminum container system. Let final temperature of the mixture be #T^@C#. Using CGS units.

Heat gained by water kept in container to change from temperature #10^@"C"# to temperature at #T^@C#. #s# for water #=1calgm^-1"^@C^-1#

#DeltaQ_"gained 1"=mst=200xx1xx(T-10)=200(T-10)#
Heat gained by aluminum container to change from temperature #10^@"C"# to temperature at #T^@C#.
#s# for aluminum #=0.215 calgm^-1"^@C^-1#

#DeltaQ_"gained 2"=mst=50xx0.215xx(T-10)=10.75(T-10)#

Total Heat gained by water and container #=DeltaQ_"gained 1"+DeltaQ_"gained 2"=200(T-10)+10.75(T-10)# #=210.75(T-10)# .....(1)

Now heat lost by steam to cool down from #100^@"C# to become water at #T^@"C"# is given by sum of heat lost in becoming water at #100^@C#, change of state, plus heat lost to become water at #T^@C#.
#L# for steam #=540calgm^-1#

#DeltaQ_"lost 1"=mL=10xx540=5400#
#DeltaQ_"lost 2"=ms(100-T)=10xx1xx(100-T)=10(100-T)#
Total heat lost by steam#=5400+10(100-T)# ......(2)

Equating (1) and (2) and solving for the required quantity
#5400+10(100-T)=210.75(T-10)#
#=>5400+1000-10T=210.75T-2107.5#
#=>220.75T=8507.5#
#=>T=38.5^@C#, rounded to one decimal place.