# An object with a mass of 4 kg, temperature of 265 ^oC, and a specific heat of 12 (KJ)/(kg*K) is dropped into a container with 39 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Jan 26, 2018

The water will not evaporate and the change in temperature is $= {60.2}^{\circ} C$

#### 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 $\Delta {T}_{o} = 265 - T$

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

The specific heat of water is ${C}_{w} = 4.186 k J k {g}^{-} 1 {K}^{-} 1$

The specific heat of the object is ${C}_{o} = 12 k J k {g}^{-} 1 {K}^{-} 1$

The mass of the object is ${m}_{0} = 4 k g$

The mass of the water is ${m}_{w} = 39 k g$

$4 \cdot 12 \cdot \left(265 - T\right) = 39 \cdot 4.186 \cdot T$

$265 - T = \frac{39 \cdot 4.186}{4 \cdot 12} \cdot T$

$265 - T = 3.4 T$

$4.4 T = 265$

$T = \frac{265}{4.4} = {60.2}^{\circ} C$

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