An object with a mass of 3 kg, temperature of 244 ^oC, and a specific heat of 37 (KJ)/(kg*K) is dropped into a container with 16 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Jun 18, 2018

Some water will evaporate.

Explanation:

The heat is transferred from the hot object to the cold water.

Let $T =$ final temperature of the object and the water

For the cold water, $\Delta {T}_{w} = T - 0 = T$

For the object $\Delta {T}_{o} = 244 - T$

${m}_{o} {C}_{o} \left(\Delta {T}_{o}\right) = {m}_{w} {C}_{w} \left(\Delta {T}_{w}\right)$

The specific heat of the object is ${C}_{o} = 37 J k {g}^{-} 1 {K}^{-} 1$

The mass of the object is ${m}_{0} = 3 k g$

The volume of water is $V = 16 L$

The density of water is $\rho = 1 k g {L}^{-} 1$

The mass of the water is ${m}_{w} = \rho V = 16 k g$

$3 \cdot 37 \cdot \left(244 - T\right) = 16 \cdot 4.186 \cdot T$

$244 - T = \frac{16 \cdot 4.186}{3 \cdot 37} \cdot T$

$244 - T = 0.603 T$

$1.603 T = 244$

$T = \frac{244}{1.603} = {152.2}^{\circ} C$

As the final temperature is $T > {100}^{\circ} C$, some water will evaporate.