# An object with a mass of 1 kg, temperature of 320 ^oC, and a specific heat of 25 J/(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?

Jan 20, 2018

For the water to evaporate, its temperature would need to reach 100ºC. Moreover, it would be necessary to take into consideration the amount of water and the energy needed to vaporize it.

First of all, we can estimate the amount of energy needed to heat the water up to 100ºC using the $Q = m c \Delta T$ relation, where $Q$ is the heat absorbed by the water (in this case) to vary the temperature of a mass $m$ of water with $c$ specific heat by $\Delta T$.

Using 1kg/L for the density of water, then our 16 L of water contain 16 kg of water. Then:

$Q = m c \Delta T$;
$Q = 16 \cancel{k g} \cdot 4 \frac{k J}{\left(\cancel{k g} \cancel{K}\right)} \cdot 100 \cancel{K}$;
$Q = 6400 k J$

We can also estimate the amount of heat needed for all this water to evaporate using the relation $Q = m L$, where $L$ is the latent heat of water to vaporize: 2264 kJ/kg.

$Q = 16 k g \cdot 2.264 \frac{k J}{k g} = 36224 k J$.

Therefore, we would need a total of $6400 + 36224 = 42624 k J$ of heat to evaporate all this water. All we have to ask ourselves now is if our hot object can, indeed, give this amount of energy to the water.

Let's check what would be the final temperature of the 1 kg object if it were to give away this amount of heat:

${Q}_{o b j e c t} = {m}_{o b j e c t} {c}_{o b j e c t} \Delta {T}_{o b j e c t}$
$42624 k J = 1 k g \cdot 0.025 k \frac{J}{k g K} \cdot \Delta T$
$\Delta T = 1.7 \cdot {10}^{6} K$;

which is a very large value and, thus, impossible for this case (meaning that the object does not posses the needed amount of heat to evaporate all the water).

Let us, then, estimate the final temperature of the ensemble. We know that on thermal equilibrium, all heat changes must sum up to zero. Then:

${Q}_{w a t e r} + {Q}_{o b j e c t} = 0$;
${m}_{w a t e r} {c}_{w a t e r} \Delta {T}_{w a t e r} + {m}_{o b j e c t} {c}_{o b j e c t} \Delta {T}_{o b j e c t} = 0$;
$16 k g \cdot 4 \frac{k J}{k g K} \cdot \left(T - 273\right) + 1 k g \cdot 0.025 \frac{k J}{k g K} \cdot \left(T - 593\right) = 0$;
$64 T - 17472 + 0.025 T - 14.825 = 0$
$64.025 T = 17486.825$
$T = 273.12 K$

Note that this is almost the same temperature at which the water started (273 K), which actually makes sense, since the specific heat of water is large and there is a considerable amount of water. On the other hand, our object has a low specific heat, yielding its high temperature drop.