# An object with a mass of 18 kg, temperature of 270 ^oC, and a specific heat of 8 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?

Jan 1, 2018

THe water will not evaporate and the change in temperature is $= {0.19}^{\circ} C$

#### 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 ${C}_{o} = 0.008 k J k {g}^{-} 1 {K}^{-} 1$

The mass of the object is ${m}_{0} = 18 k g$

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

$18 \cdot 0.008 \cdot \left(270 - T\right) = 48 \cdot 4.186 \cdot T$

$270 - T = \frac{48 \cdot 4.186}{18 \cdot 0.008} \cdot T$

$270 - T = 1395.3 T$

$1396.3 T = 270$

$T = \frac{270}{1396.3} = {0.19}^{\circ} C$

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