# An object with a mass of 4 kg, temperature of 341 ^oC, and a specific heat of 12 (KJ)/(kg*K) is dropped into a container with 42 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Mar 6, 2017

The water does not evaporate and the temperature changes by 73.1ºC

#### 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 $\Delta {T}_{o} = 341 - T$

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

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

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

${m}_{0} {C}_{o} \cdot \left(341 - T\right) = {m}_{w} \cdot 4.186 \cdot T$

$4 \cdot 12 \cdot \left(341 - T\right) = 42 \cdot 4.186 \cdot T$

$341 - T = \frac{42 \cdot 4.186}{48} \cdot T$

$341 - T = 3.66 T$

$4.66 T = 341$

T=341/4.66=73.1ºC

The water does not evaporate