An object with a mass of 16 kg, temperature of 180 ^oC, and a specific heat of 5 J/(kg*K) is dropped into a container with 48 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Mar 21, 2018

The water does not evaporate and the change in the tempertaure is $= {0.07}^{\circ} C$

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 $\Delta {T}_{o} = 180 - T$

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

The specific heat of the object is ${C}_{o} = 0.005 k J k {g}^{-} 1 {K}^{-} 1$

The mass of the object is ${m}_{0} = 16 k g$

The volume of water is $V = 48 L$

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

The mass of the water is ${m}_{w} = \rho V = 48 k g$

$16 \cdot 0.005 \cdot \left(180 - T\right) = 48 \cdot 4.186 \cdot T$

$180 - T = \frac{48 \cdot 4.186}{16 \cdot 0.005} \cdot T$

$180 - T = 2511.6 T$

$2512.6 = 180$

$T = \frac{180}{2512.6} = {0.07}^{\circ} C$

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 not very high and the specific heat is low.