# An object with a mass of 12 kg, temperature of 210 ^oC, and a specific heat of 7 J/(kg*K) is dropped into a container with 32 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Jul 18, 2017

$\Delta T = {208.8}^{\circ} C$ The water will evaporate.

#### Explanation:

${210}^{\circ} C \approx 483 K$
${0}^{\circ} C \approx 273 K$
$32 L = 0.032 {m}^{3} \cong 0.032 k g$

${m}_{o} {C}_{o} \Delta {T}_{o} = {m}_{w} {C}_{w} \Delta {T}_{w}$

$12 \cdot 7 \cdot \left(483 - T\right) = 0.032 \cdot 4.18 \cdot \left(273 + T\right)$

#40572-84T=36.51648+0.13376T

$84.13376 T = 40535.4835$

$T = \frac{40535.4835}{84.13376} = 481.798074 K = {208.798074}^{\circ} C \approx {208.8}^{\circ} C$

$0 + 208.8 = {208.8}^{\circ} C$

As $208.8 > 100$ the water will evaporate.