# An object with a mass of 18 kg, temperature of 240 ^oC, and a specific heat of 8 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?

Apr 17, 2017

The water does not evaporate and the change in temperature is =0.17ºC

#### 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 $\Delta {T}_{o} = 240 - T$

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

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

${C}_{o} = 0.008 k J k {g}^{-} 1 {K}^{-} 1$

$18 \cdot 0.008 \cdot \left(240 - T\right) = 48 \cdot 4.186 \cdot T$

$240 - T = \frac{48 \cdot 4.186}{18 \cdot 0.008} \cdot T$

$240 - T = 1395.3 T$

$1396.3 T = 240$

T=240/1396.3=0.17ºC

As T<100ºC, the water does not evaporate.