# An object with a mass of 16 kg, temperature of 270 ^oC, and a specific heat of 5 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?

##### 1 Answer
Dec 29, 2017

The water does not evaporate and the change in temperature is $= {0.26}^{\circ} C$

#### 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 $\Delta {T}_{o} = 270 - T$

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

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

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

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

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

$16 \cdot 0.005 \cdot \left(270 - T\right) = 32 \cdot 4.186 \cdot T$

$270 - T = \frac{32 \cdot 4.186}{16 \cdot 0.005} \cdot T$

$270 - T = 1674.4 T$

$1675.4 T = 270$

$T = \frac{270}{1675.4} = {0.16}^{\circ} C$

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