An object with a mass of 2 kg, temperature of 311 ^oC, and a specific heat of 18 (KJ)/(kg*K) is dropped into a container with 37 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Sep 16, 2017

The water does not evaporate and the change in temperature is $= {58.7}^{\circ} C$

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 $\Delta {T}_{o} = 311 - T$

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

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

The specific heat of the object is ${C}_{o} = 18 k J k {g}^{-} 1 {K}^{-} 1$

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

The mass of the water is ${m}_{w} = 37 k g$

$2 \cdot 18 \cdot \left(311 - T\right) = 37 \cdot 4.186 \cdot T$

$311 - T = \frac{37 \cdot 4.186}{2 \cdot 18} \cdot T$

$311 - T = 4.30 T$

$5.30 T = 311$

$T = \frac{311}{5.30} = {58.7}^{\circ} C$

As $T < {100}^{\circ} C$, the water does not evaporate