# An object with a mass of 3 kg, temperature of 184 ^oC, and a specific heat of 37 (KJ)/(kg*K) is dropped into a container with 15 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Jan 15, 2018

The water will evaporate

#### 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 $\Delta {T}_{o} = 184 - T$

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

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

The specific heat of the object is ${C}_{o} = 37 k J k {g}^{-} 1 {K}^{-} 1$

The mass of the object is ${m}_{0} = 3 k g$

The mass of the water is ${m}_{w} = 15 k g$

$3 \cdot 37 \cdot \left(184 - T\right) = 15 \cdot 4.186 \cdot T$

$184 - T = \frac{15 \cdot 4.186}{3 \cdot 37} \cdot T$

$184 - T = 0.566 T$

$1.566 T = 184$

$T = \frac{184}{1.566} = {117.5}^{\circ} C$

As $T > {100}^{\circ} C$, the water will evaporate