# An object with a mass of 18 kg, temperature of 240 ^oC, and a specific heat of 6 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?

Feb 24, 2017

The water does not evaporate and the rise in temperature ot the water is =0.13ºC

#### Explanation:

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

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

For the object $\Delta {T}_{o} = 240 - T$

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

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

${C}_{o} = 6 J k {g}^{-} 1 {K}^{-} 1$

${m}_{0} {C}_{o} \cdot \left(240 - T\right) = {m}_{w} \cdot 4186 \cdot T$

$18 \cdot 6 \cdot \left(240 - T\right) = 48 \cdot 4186 \cdot T$

$240 - T = \frac{48 \cdot 4186}{108} \cdot T$

$240 - T = 1860.4 T$

$1861.4 T = 240$

T=240/1861.4=0.13ºC

The water does not evaporate