# An object with a mass of 6 kg, temperature of 50 ^oC, and a specific heat of 4 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?

Mar 3, 2018

The water will not evaporate and the change in temperature is $= {0.012}^{\circ} C$

#### 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 $\Delta {T}_{o} = 50 - T$

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

The [
The specific heat of the object is ${C}_{o} = 0.004 k J k {g}^{-} 1 {K}^{-} 1$

The mass of the object is ${m}_{0} = 6 k g$

The volume of water is $V = 24 L$

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

The mass of the water is ${m}_{w} = \rho V = 24 k g$

$6 \cdot 0.004 \cdot \left(50 - T\right) = 24 \cdot 4.186 \cdot T$

$50 - T = \frac{24 \cdot 4.186}{6 \cdot 0.004} \cdot T$

$50 - T = 4186 T$

$4187 T = 50$

$T = \frac{50}{4187} = {0.012}^{\circ} C$

As the final temperature is $T < {100}^{\circ} C$, the water will not evaporate. We expect this result as the temperature of the object is $< {100}^{\circ} C$ and the specific heat is low.