Enthalpy

Heating Curve

When heat power is supplied to a hard (like ice) at a steady rate by method of an electrical heating coil, we discover that the temperature climbs steadily until the melting allude is reached and the an initial signs of liquid formation end up being evident. Thereafter, even though we room still offering heat power to the system, the temperature remains consistent as long as both liquid and solid room present. ~ above the graph below, this is represented by the flat line, where energy is being added to the ice, yet no readjust is emerging in the temperature. Every energy added to the device at this phase is offered to transform solid ice into liquid water.

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Once all of the sample is in the liquid phase, the addition of power now increases the temperature until the boiling suggest is reached and also the very first signs that gas formation are seen. The temperature remains continuous even though power is being added to the system. The power is being used to transform the fluid to a gas. Once all of the sample is in the gas phase, added energy deserve to be added to increase the temperature that the gas.

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Figure (PageIndex1): heater curve the water at 1 atm. By community College Consortium for Bioscience Credentials (Own work) , via Wikimedia Commons

This macroscopic behavior demonstrates quite clearly that energy must be supplied to a heavy in order come melt or vaporize it. ~ above a microscopic level melting or vaporization entails separating molecules which room attracted to every other. The lot of energy needed to separate the molecule is proportional to the intermolecular forces between the molecules.



Example (PageIndex1): heat Energy

Calculate the heat power required to (a) vaporize 100. G of lead, (b) melt 100. G the lead, (c) vaporize 100. G water, and (d) melt 100. G of water.

Solution

(a)To vaporize 100. G of lead:

< extPb(l) ightarrow extPb(g);;;;;;;;;; riangle H_vap = 178 dfrac kJmol>

<100.;g imesdfrac 1;mol; extPb207.2;g; extPb imes dfrac178;kJmol = 85.9; kJ>

(b) come melt 100. G that lead:

< extPb(s) ightarrow extPb(l);;;;;;;;;; riangle H_fus = 4.77 dfrac kJmol>

<100.;g imesdfrac 1;mol; extPb207.2;g; extPb imes dfrac4.77;kJmol = 2.30; kJ>

(c) to vaporize 100. G that water:

< extH_2 extO(s) ightarrow extH_2 extO(l);;;;;;;;;; riangle H_vap =40.657dfrac kJmol>

<100.;g imesdfrac 1;mol; extH_2 extO18.0;g; extH_2 extO imes dfrac40.657;kJmol = 226; kJ>

(d) come melt 100. G that water:

< extH_2 extO(s) ightarrow extH_2 extO(l);;;;;;;;;; riangle H_fus = 6.01 dfrac kJmol>

<100.;g imesdfrac 1;mol; extH_2 extO18.0;g; extH_2 extO imes dfrac6.01;kJmol = 33.4; kJ>

It could be surprising that the heat compelled to melt or vaporize 100 g of command is therefore much less than that need to melt or vaporized water. First, the temperature at which the problem melts has nothing to execute with the enthalpy that fusion. Remember, us are just looking in ~ the energy required to adjust the phase, no the power required to gain the problem to the melting or boil point.

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Also keep in mind that the enthalpies of blend and vaporization are provided as kJ per mole. Although the water and also the lead have actually the exact same mass, the mole of each substance is really different (5.5 mole of water vs. 0.48 mole of lead).