# 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?

Mar 28, 2016

Rise in the temperature of water$= {52.1}^{\circ} \text{C}$ 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
$\Delta Q = m s t$,
where $m , s \mathmr{and} t$ are the mass, specific heat and rise or gain in temperature of the object.

Also from Law of conservation of energy:
$\Delta {Q}_{\text{lost"=Delta Q_"gained}}$

In the given problem heat is lost by object and gained by water. Let the final temperatuer of the mixture be ${t}^{\circ} \text{C}$

Heat lost by object to cool down from ${270}^{\circ} \text{C to " t^@"C}$ is given by
$\Delta {Q}_{\text{lost"=2}} \cdot 18 \cdot \left(270 - t\right)$
$= 36 \left(270 - t\right)$ ........(1)

Heat gained by water to change from ${0}^{\circ} \text{C}$ to ${t}^{\circ} \text{C}$ is given by
$\Delta {Q}_{\text{gained}} = m s t$
$\Delta {Q}_{\text{gained}} = 36 \times 4.1813 \times \left(t - 0\right)$
$\Delta {Q}_{\text{gained}} = 150.5268 \left(t - 0\right)$......(2)
Specific heat of water is $4.1813 J k {g}^{-} 1 {K}^{-} 1$ and mass of 1 liter of water is $1 k g$.
Equating (1) and (2) and solving for the required quantity
$150.5268 t = 36 \left(270 - t\right)$
$150.5268 t = 9720 - 36 t$
or $186.5268 t = 9720$
or $t = {52.1}^{\circ} \text{C}$ rounded to one decimal place.