# An object with a mass of 2 kg, temperature of 50 ^oC, and a specific heat of 19 J/(kg*K) is dropped into a container with 24 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Feb 28, 2017

The water does not evaporate. The change in temperature is =0.16ºC

#### Explanation:

Let $T =$ final temperature of the object and the water

For the cold water, $\Delta {T}_{w} = T - 0 = T$

For the object $\Delta {T}_{o} = 50 - T$

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

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

${C}_{o} = 19 J k {g}^{-} 1 {K}^{-} 1$

${m}_{0} {C}_{o} \cdot \left(50 - T\right) = {m}_{w} \cdot 4186 \cdot T$

$2 \cdot 19 \cdot \left(50 - T\right) = 24 \cdot 4186 \cdot T$

$50 - T = \frac{24 \cdot 4186}{38} \cdot T$

$50 - T = 306.9 T$

$307.9 T = 50$

T=50/307.9=0.16ºC

The water does not evaporate