# An object with a mass of 1 kg, temperature of 120 ^oC, and a specific heat of 32 J/(kg*K) is dropped into a container with 12 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Apr 18, 2018

The water will not evaporate and the change in temperature is $= {0.08}^{\circ} C$

#### 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 $\Delta {T}_{o} = 120 - T$

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

The specific heat of the object is ${C}_{o} = 0.032 k J k {g}^{-} 1 {K}^{-} 1$

The mass of the object is ${m}_{0} = 1 k g$

The volume of water is $V = 12 L$

The density of water is $\rho = 1 k g {L}^{-} 1$

The mass of the water is ${m}_{w} = \rho V = 12 k g$

$1 \cdot 0.032 \cdot \left(120 - T\right) = 12 \cdot 4.186 \cdot T$

$120 - T = \frac{12 \cdot 4.186}{1 \cdot 0.032} \cdot T$

$120 - T = 1569.8 T$

$1570.8 T = 120$

$T = \frac{120}{1570.8} = {0.08}^{\circ} C$

As the final temperature is $T < {100}^{\circ} C$, the water will not evaporate.