# An object with a mass of 3 kg, temperature of 145 ^oC, and a specific heat of 23 J/(kg*K) is dropped into a container with 32 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Dec 4, 2017

The water does not evaporate and the change in temperature 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 $\Delta {T}_{o} = 145 - T$

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

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

The specific heat of the object is ${C}_{o} = 0.023 k J k {g}^{-} 1 {K}^{-} 1$

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

The mass of the water is ${m}_{w} = 32 k g$

$3 \cdot 0.023 \cdot \left(145 - T\right) = 32 \cdot 4.186 \cdot T$

$145 - T = \frac{32 \cdot 4.186}{3 \cdot 0.023} \cdot T$

$145 - T = 1941.3 T$

$1942.3 T = 145$

$T = \frac{145}{1942.3} = {0.07}^{\circ} C$

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

You would expect this result, as the mass of the object and its specfic heat are small compared to the mass of water and the specific heat of water.