# An object with a mass of 2 kg, temperature of 214 ^oC, and a specific heat of 12 (KJ)/(kg*K) is dropped into a container with 25 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

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Mar 28, 2018

#### Answer:

The water will not evaporate and the change in temperature is $= {39.9}^{\circ} C$

#### 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 $\Delta {T}_{o} = 214 - T$

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

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

The mass of the object is ${m}_{0} = 2 k g$

The volume of water is $V = 25 L$

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

The mass of the water is ${m}_{w} = \rho V = 25 k g$

$2 \cdot 12 \cdot \left(214 - T\right) = 25 \cdot 4.186 \cdot T$

$214 - T = \frac{25 \cdot 4.186}{2 \cdot 12} \cdot T$

$214 - T = 4.36 T$

$5.36 = 214$

$T = \frac{214}{5.36} = {39.9}^{\circ} C$

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

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