Draw a heating curve (such as the one in Figure 11.36 ) for...

Draw a heating curve (such as the one in Figure $11.36$ ) for 1 mole of benzene beginning at $0{ }^{\circ} \mathrm{C}$ and ending at $100{ }^{\circ} \mathrm{C}$. Assume that the values given here are constant over the relevant temperature ranges. $$ \begin{array}{ll} \hline \text { Melting point } & 5.4^{\circ} \mathrm{C} \\ \hline \text { Boiling point } & 90.1^{\circ} \mathrm{C} \\ \hline \Delta \mathrm{H}_{\text {fus }} & 9.9 \mathrm{k} / \mathrm{mol} \\ \hline \Delta \mathrm{H}_{\text {vap }} & 30.7 \mathrm{k} / \mathrm{mol} \\ \hline \mathrm{C}_{\text {s }, \text { solid }} & 118 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\ \hline \mathrm{C}_{\text {s }, \text { liquid }} & 135 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\ \hline \mathrm{C}_{\text {s, gas }} & 104 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\ \hline \end{array} $$

Draw a heating curve (such as the one in Figure $11.36$ ) for 1 mole of benzene beginning at $0{ }^{\circ} \mathrm{C}$ and ending at $100{ }^{\circ} \mathrm{C}$. Assume that the values given here are constant over the relevant temperature ranges.
$$
\begin{array}{ll}
\hline \text { Melting point } & 5.4^{\circ} \mathrm{C} \\
\hline \text { Boiling point } & 90.1^{\circ} \mathrm{C} \\
\hline \Delta \mathrm{H}_{\text {fus }} & 9.9 \mathrm{k} / \mathrm{mol} \\
\hline \Delta \mathrm{H}_{\text {vap }} & 30.7 \mathrm{k} / \mathrm{mol} \\
\hline \mathrm{C}_{\text {s }, \text { solid }} & 118 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\
\hline \mathrm{C}_{\text {s }, \text { liquid }} & 135 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\
\hline \mathrm{C}_{\text {s, gas }} & 104 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\
\hline
\end{array}
$$

Key Concepts

Heating Curve

A heating curve is a graphical representation that shows how a substance’s temperature changes as heat is added to it. The curve typically consists of segments where the temperature increases with added energy and flat portions where phase changes occur. During the flat regions, energy is absorbed for the phase transitions without a change in temperature, while in sloped regions, all the added heat increases the temperature according to the substance's specific heat capacity.

Phase Transitions

Phase transitions are processes in which a substance changes its state of matter, such as melting (solid to liquid) and boiling (liquid to gas). These transitions occur at constant temperature and require specific amounts of energy known as latent heats. In a heating curve, the phase transitions appear as horizontal segments because the heat energy is used to overcome the intermolecular forces rather than to increase temperature.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to change the temperature of a unit amount of a substance by one degree. It defines the slopes of the regions on a heating curve where the substance’s temperature is rising without a phase change. Each phase (solid, liquid, gas) has its own specific heat capacity, which influences how rapidly the temperature increases with added heat.

Latent Heat

Latent heat is the amount of energy required for a substance to undergo a phase change without a change in temperature. This includes latent heat of fusion and vaporization, which are the energies needed for melting and boiling, respectively. In a heating curve, the latent heat is represented by the flat portions where energy is absorbed but the temperature remains constant during the phase transition.

Transcript

00:01 In this question we're going to be looking into the heating cave of benzene.

00:06 And what we want here is a t against q diagram for heating benzene from 0 to 100 degrees celsius.

00:14 So within this temperature change, there are a lot of heat changes that okay and phase changes.

00:20 So for example, the first set that we're going to look at is the change in temperature from 0 degree celsius to 5 .4 degrees celsius, which is the melting temperature, which is the melting point.

00:31 Of benzene.

00:32 So generally we know that our heat is equal to q is equal to n the specific capacity multiplied by the temperature change.

00:44 So our q is equal to 1 more of benzene multiplied by the specific heat capacity of solid benzene which is equal to 118 multiplied by the temperature change which is 5 .4 kelvin.

00:56 Remember the scale of the degree celsius and the kelvin is the same.

01:01 So the change is the same so our q this is equal to 637 .2 and this is in joules to convert this to kilojoles this is going to be 0 .637 kilojoles so next now that we have reached melting point we have to melt our benzene and this occurs at constant temperature we are changing the phase and this process occurs at constant temperature we are changing the phase and this process occurs at constant and temperature so our cue is going to be n multiplied by the end of change of fusion and this is equal to 1 multiplied by 9 .9 kilojoules per more so our cue here is going to be equal to 9 .9 kilojols now that we have a liquid benzene we have to hit it right up to its boiling point which is 9 .0 .1 so we are now moving from 5 .0 point 4 degrees celsius up to 90 .1 degrees celsius.

02:06 So our cue here is going to be n multiplied, which is 1, multiplied by the specific capacity of liquid benzene, which is 135, multiplied by the temperature change, which is 8 .4 .7.

02:22 This is 8 .47 degrees celsius, which is also 8 .7 kelvin.

02:27 So our cue here is going to be equal to 11434 .3 .7.

02:32 Joles and this is equal to 11 .435 kilojoules.

02:38 Now that we are at the boiling point we have to change the phase that is to boil and this occurs at a constant temperature.

02:46 We are changing phase at constant temperature of 90 .1.

02:50 So our q is equal to one more multiplied by the enthalpy change of vaporization of liquid benzene which is 30 .7 kilojoules per more.

03:01 So our q is equal to 30 .7 kilojoules.

03:05 Now that we have our paper, we can then heat that paper up to 100 degrees celsius.

03:12 So now the change in temperature here is going to be from 90 .1 up to 100 degrees celsius.

03:18 Looking at this, our q is going to be equal to number of moles multiplied by the specific heat capacity...

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