# An object with a mass of 2 kg, temperature of 150 ^oC, and a specific heat of 24 J/(kg*K) is dropped into a container with 18 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Mar 4, 2017

The water does not evaporate and the change in temperature is 0.01ºC

#### 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 $\Delta {T}_{o} = 150 - T$

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

${C}_{w} = 4186 J k {g}^{-} 1 {K}^{-} 1$

${C}_{o} = 24 J k {g}^{-} 1 {K}^{-} 1$

${m}_{0} {C}_{o} \cdot \left(150 - T\right) = {m}_{w} \cdot 4186 \cdot T$

$2 \cdot 24 \cdot \left(150 - T\right) = 18 \cdot 4186 \cdot T$

$150 - T = \frac{18 \cdot 4186}{48} \cdot T$

$150 - T = 1569.75 T$

$1570.75 T = 150$

T=150/1570.75=0.01ºC

The water does not evaporate