# An object with a mass of 6 kg, temperature of 315 ^oC, and a specific heat of 12 (KJ)/(kg*K) is dropped into a container with 39 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Jul 25, 2017

The water does not evaporate and the change in temperature is $= {96.4}^{\circ} C$

#### 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 $\Delta {T}_{o} = 315 - T$

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

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

${C}_{o} = 12 k J k {g}^{-} 1 {K}^{-} 1$

$6 \cdot 12 \cdot \left(315 - T\right) = 39 \cdot 4.186 \cdot T$

$315 - T = \frac{39 \cdot 4.186}{6 \cdot 12} \cdot T$

$315 - T = 2.267 T$

$3.267 T = 315$

$T = \frac{315}{3.267} = {96.4}^{\circ} C$

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