An object with a mass of 6 kg, temperature of 160 ^oC, and a specific heat of 9 J/(kg*K) is dropped into a container with 27 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Feb 27, 2017

The water does not evaporate, the final temperature is =0.1ºC

Explanation:

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

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

For the object $\Delta {T}_{o} = 160 - T$

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

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

${C}_{o} = 9 J k {g}^{-} 1 {K}^{-} 1$

${m}_{0} {C}_{o} \cdot \left(160 - T\right) = {m}_{w} \cdot 4186 \cdot T$

$6 \cdot 9 \cdot \left(160 - T\right) = 27 \cdot 4186 \cdot T$

$160 - T = \frac{27 \cdot 4186}{54} \cdot T$

$160 - T = 2093 T$

$2094 T = 160$

T=160/2094=0.1ºC

The water does not evaporate