Question

A $65c{m}^3$ block of iron is removed from an${800}^{0}C$ furnace and

immediately dropped into 200 mL of ${20}^0$C water. What fraction

of the water boils away?

Short answer

The iron block in this example provided energy to the water until it reached its phase shift. The leftover energy then converted a portion of the water to steam until the system reached equilibrium.

Step-by-step solution

Given information

Given:

We have $65{\mathrm{cm}}^3$of iron at $800°\mathrm{C},200\mathrm{mL}$of water at $20°\mathrm{C}$.

Formula used:

role="math" localid="1650517403634" $m_{\text{iron}}\Delta C_{p,\text{iron}}\Delta T_{i,\text{iron}}+m_{\text{water}}\Delta C_{p,\text{water}}\Delta T_{i,\text{water}}=m_{\text{iron}}\Delta C_{p,\text{iron}}\Delta T_f+m_{\text{water}}\Delta C_{p,\text{water}}\Delta T_f+m_{\text{steam}}\Delta C_{\text{steam}}$

Simplification

Let us solve the$m_{\text{steam}}$

$m_{\text{iron}}\Delta C_{p,\text{iron}}\Delta T_{i,\text{iron}}+m_{\text{water}}\Delta C_{p,\text{water}}\Delta T_{i,\text{water}}-m_{\text{iron}}\Delta C_{p,\text{iron}}\Delta T_f-m_{\text{water}}\Delta C_{\text{water}}\Delta T_f=m_{\text{steam}}\Delta C_{p\text{steam}}$

$$\begin{gathered} m_{\text{iron}}=65\left[{\mathrm{cm}}^3\right]\Delta 7.874\left[\frac{\mathrm{g}}{{\mathrm{cm}}^3}\right]=511.81[\mathrm{g}] \\ {m}_{\text{water}}=200\left[mL\right]\Delta 1\left[\frac{g}{mL}\right]=200\left[g\right] \\ {C}_{p,\text{water}}=4.181\left[\frac{J}{g\Delta K}\right] \\ {C}_{p,\text{iron}}=0.450\left[\frac{J}{g\Delta K}\right] \end{gathered}$$

Latent heat to vaporize water:

$$\begin{gathered} C_{\text{steam}}=2257\left[\frac{\mathrm{J}}{\mathrm{g}}\right] \\ {T}_{i,\text{water}}=20.0\left[{}^{\circ }\mathrm{C}\right]+273.15=293.15\left[K\right] \\ {T}_{i,\text{iron}}=800\left[{}^{\circ }\mathrm{C}\right]+273.15=1073.15\left[K\right] \\ {T}_f=100.0\left[{}^{\circ }\mathrm{C}\right]+273.15=373.15\left[K\right] \end{gathered}$$

$$\begin{gathered} m_{\text{steam}}=\frac{511.81[\mathrm{g}]\Delta 0.45\left[\frac{\mathrm{J}}{\mathrm{g}AK}\right]\Delta (1073.15[K]-373.15[K\left]\right)+200|\mathrm{g}|\Delta 4.181\left[\frac{\mathrm{J}}{\mathrm{g}AK}\right]\Delta (293.15[K]-373.15[K\left]\right)}{2257\left[\frac{\mathrm{J}}{\mathrm{K}}\right]} \\ {m}_{\text{steam}}=41.79[\mathrm{g}] \\ x=\frac{m_{\text{staia}}}{m_{\text{vatir}}}=\frac{41.79\mid \mathrm{g}]}{200\mid \mathrm{g}]}=0.209=20.9\% \end{gathered}$$