Question

An object with a mass of `2 kg`, temperature of `270 ^oC`, and a specific heat of `18 J/(kg*K)` is dropped into a container with `36 L` of water at `0^oC`. Does the water evaporate? If not, by how much does the water's temperature change?

Answer 1

Rise in the temperature of water`=52.1^@"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`,
where `m,s and t` are the mass, specific heat and rise or gain in temperature of the object.

Also from Law of conservation of energy:
`Delta Q_"lost"=Delta Q_"gained"`

In the given problem heat is lost by object and gained by water. Let the final temperatuer of the mixture be `t^@"C"`

Heat lost by object to cool down from `270^@"C to " t^@"C"` is given by
`Delta Q_"lost"=2"cdot18cdot (270-t)`
`=36 (270-t)` ........(1)

Heat gained by water to change from `0^@"C"` to `t^@"C"` is given by
`DeltaQ_"gained"=mst`
`DeltaQ_"gained"=36xx4.1813xx(t-0)`
`DeltaQ_"gained"=150.5268(t-0)`......(2)
Specific heat of water is `4.1813Jkg^-1K^-1` and mass of 1 liter of water is `1kg`.
Equating (1) and (2) and solving for the required quantity
`150.5268t=36 (270-t)`
`150.5268t=9720-36t`
or `186.5268t=9720`
or `t=52.1^@"C"` rounded to one decimal place.