# Molten iron is extremely hot, averaging about 1,500 C. The specific heat of iron is 0.46 J/gC. How much heat is released to the atmosphere when 1 kg molten iron cools to room temperature (25 C)?

Jun 24, 2017

900 Kj

#### Explanation:

Since the type of iron is not specified then it is assumed to be 'Cast Iron' which has a melting point of 1204 deg-Celsius*.
*http://www.onlinemetals.com/meltpt.cfm

The total heat transfer would be ,,,

Q_"Total = "Sigma (Q_("molten") + Q_("freezing") + Q_("cooling"))
..............................................................................................................................
=> ${Q}_{\text{molten"=(mcDeltaT)_"molten}}$
= $\left(1000 g \times 0.18 {\text{J/g}}^{o} C \times {\left(1500 - 1204\right)}^{o} C\right) = 53 , 280$ Joules

(Specific Heat of Molten Iron) http://www.engineeringtoolbox.com/liquid-metal-boiling-points-specific-heat-d_1893.html
..............................................................................................................................
=> ${Q}_{\text{freezing"=(mDeltaH_f)_"freezing}}$
= $\left(1000 g \times 272 \frac{J}{g}\right) = 272 , 000$ Joules

(Heat of Fusion of Molten Iron) http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html

...............................................................................................................................
=> ${Q}_{\text{cooling"=(mcDeltaT)_"cooling"to""25^oC}}$
= $\left(1000 g \times 0.46 \frac{J}{g} ^ o C \times {\left(1204 - 25\right)}^{o} C\right)$ = 542,340 Joules

(Specific Heat of Iron (s) = 0.46 $\frac{J}{g} ^ o C$ as given in problem data)
..............................................................................................................................
${Q}_{\text{Total}} = \left(53 , 280 + 272 , 000 + 542 , 340\right) J$ = 867,620 Joules
$\approx 9 \times {10}^{5} \text{Joules}$ = $900 K j$

Jun 24, 2017

I got $\text{1020 kJ}$ were RELEASED into the atmosphere, ignoring phase changes between the $\alpha$, $\delta$, and $\gamma$ phases and just looking at the temperature changes.

You can get more context here:
https://en.wikipedia.org/wiki/IronPhase_diagram_and_allotropes

and you can examine the specific heat capacity variations more closely here:

On another note, this $\text{1020 kJ}$ is quite a bit higher than what one would normally expect to get, $\text{655.5 kJ}$, due to taking into account the huge variation in heat capacity across ${1475}^{\circ} \text{C}$.

If you simply assume a ${C}_{P}$ of $\text{0.46 J/g"cdot"K}$ throughout, you would get $\text{655.5 kJ}$ instead ($656$ to three sig figs).

There is a HUGE assumption here that iron's specific heat capacity doesn't change from ${25}^{\circ} \text{C}$ to ${1500}^{\circ} \text{C}$, which is clearly not true. Here is the phase diagram of iron:

Since all these phases at $\text{1 bar}$ are solids, we are safe in assuming there is no major enthalpy of solid-solid phase transitions to worry about.

However, the specific heat capacity ${C}_{P}$ at constant pressure changes drastically as we transition through $\alpha$, $\gamma$, and $\delta$ phases:

[

The wonky curve is the $\alpha$ and $\delta$ phase, and the linear curve is the $\gamma$ phase. Here's how I would treat this:

• $\alpha$-phase, from $\text{298.15 K}$ up to $\text{700 K}$ (${426.85}^{\circ} \text{C}$), using an average of ${C}_{P} \approx \text{29.656 J/mol"cdot"K}$ (at $\approx \text{500 K}$), or $\text{0.531 J/g"cdot"K}$.
• $\alpha$-phase, from $\text{700 K}$ to $\text{935 K}$ (${661.85}^{\circ} \text{C}$) using an average of ${C}_{P} \approx \text{40.149 J/mol"cdot"K}$ (at $\approx \text{816 K}$), or $\text{0.719 J/g"cdot"K}$
• $\alpha$-phase, from $\text{935 K}$ to $\text{1042 K}$ (${768.85}^{\circ} \text{C}$) using an average of ${C}_{P} \approx \text{59.442 J/mol"cdot"K}$ (at $\approx \text{1010 K}$), or $\text{1.064 J/g"cdot"K}$
• $\alpha$-phase, from $\text{1042 K}$ to $\text{1100 K}$ (${826.85}^{\circ} \text{C}$) using an average of ${C}_{P} \approx \text{65.743 J/mol"cdot"K}$ (at $\approx \text{1068 K}$), or $\text{1.177 J/g"cdot"K}$
• $\alpha$-phase, from $\text{1100 K}$ to $\text{1183.15 K}$ (${910}^{\circ} \text{C}$, the $\alpha \to \gamma$ transition temperature) using an average of ${C}_{P} \approx \text{43.029 J/mol"cdot"K}$ (at $\approx \text{1150 K}$), or $\text{0.770 J/g"cdot"K}$
• $\gamma$-phase, from $\text{1183.15 K}$ to $\text{1667.15 K}$ (${1394}^{\circ} \text{C}$, the $\gamma \to \delta$ transition temperature) using an average of ${C}_{P} \approx \text{35.856 J/mol"cdot"K}$ (at $\approx \text{1420 K}$), or $\text{0.642 J/g"cdot"K}$
• $\delta$-phase, from $\text{1667.15 K}$ to $\text{1773.15 K}$ (${1500}^{\circ} \text{C}$!), using an average of ${C}_{P} \approx \text{41.764 J/mol"cdot"K}$ (at $\approx \text{1722 K}$), or $\text{0.748 J/g"cdot"K}$.

Aren't you glad we aren't doing phase changes? :-)

So, we would have the heat of cooling as the negative of the heat of heating:

${q}_{\text{cool}} = - \left({q}_{1} + . . . + {q}_{7}\right)$

$= - m \left({C}_{P 1} \Delta {T}_{0 \to 1} + . . . + {C}_{P 7} \Delta {T}_{6 \to 7}\right)$

I'll leave the units out, but you know that they are $\text{J/g"cdot"K}$ for ${C}_{P}$ and $\text{K}$ for $T$. The mass is in $\text{g}$.

$= - 1000 \cdot \left[0.531 \left(700 - 298.15\right) + 0.719 \left(935 - 700\right) + 1.064 \left(1042 - 935\right) + 1.177 \left(1100 - 1042\right) + 0.770 \left(1183.15 - 1100\right) + 0.642 \left(1667.15 - 1183.15\right) + 0.748 \left(1773.15 - 1667.15\right)\right]$

Each phase then approximately contributes:

$= \overbrace{- \text{628487 J")^(alpha" phase") + overbrace(-"310728 J")^(gamma" phase") + overbrace(-"79288 J")^(delta" phase}}$

$\approx$ $- 1.020 \times {10}^{6}$ $\text{J}$,

or about $\textcolor{b l u e}{- \text{1020 kJ}}$, to three sig figs.

Jun 24, 2017

Thermal history of cooling cast iron

#### Explanation:

Thermal history of cooling cast iron from ${1500}^{o} C$ to ${25}^{o} C$ ...