# An object with a mass of 2 kg, temperature of 150 ^oC, and a specific heat of 24 J/(kg*K) is dropped into a container with 15 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Aug 19, 2017

The water doesn't evaporate

#### Explanation:

$V = 15 L = 0.015 {m}^{- 3}$
$\rho = 1000 k g \cdot {m}^{- 3}$

$\rho = \frac{m}{V}$

$m = 0.015 \cdot 1000$
$\Rightarrow m = 15 k g$

Let final temperature of object and water be ${T}_{f}$

According to the principle of calorimetry,

${Q}_{o} = {Q}_{w}$
$2 \cdot 24 \cdot \left(150 - {T}_{f}\right) = 15 \cdot 4.2 \cdot {T}_{f}$
$\Rightarrow {T}_{f} \approx {64.8}^{o} C$

$\therefore$ The water wouldn't evaporate and $\Delta T \approx {65}^{o} C$