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?

1 Answer
Mar 28, 2016

Answer:

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.