Question

Question #f09ab

Answer 1

`38.5^@C`, rounded to one decimal place.

Explanation:

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.