# 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 48 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Sep 7, 2017

The water does not evaporate and the change in temperature is negligible.

#### 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 $\Delta {T}_{o} = 210 - T$

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

${C}_{w} = 4.186 k J k {g}^{-} 1 {K}^{-} 1$

${C}_{o} = 0.007 k J k {g}^{-} 1 {K}^{-} 1$

${m}_{0} = 12 k g$

$12 \cdot 0.007 \cdot \left(210 - T\right) = 48 \cdot 4.186 \cdot T$

$210 - T = \frac{48 \cdot 4.186}{12 \cdot 0.007} \cdot T$

$210 - T = 2392 T$

$2393 T = 210$

$T = \frac{210}{2393} = {0.00004}^{\circ} C$

As $T < {100}^{\circ} C$, the water does not evaporate