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
A08
Mar 28, 2016

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.

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