# Intro # Heat Capacity $Q = mc \Delta T$ # EXTRA: Specific Latent Heat When something melts, condenses, changes state, energy, instead of translating to the kinetic energy of the particles in the 'something' goes into breaking the bonds between atoms. %%diagram!%% The rate to which this energy transfer happens is actually proportional to the mass of the sample; we call the constant the **specific latent heat** of the sample, $\ell$. So the formula for the energy required to melt a full sample would be: $E = m\ell$ Where: - $m$ **is the initial mass of the sample** - **$\ell$** **is the specific latent heat of the sample** - $E$ **is the energy put/removed into the sample from its surroundings** # EXTENSION: Molar Specific Heats Remember the equation for the amount of heat energy transferred? Well, we're going to go back to that, this time incorporating our knowledge of the different thermodynamic systems. %%GIVE A LINK TO THE SYSTEMS!%% Let's start with a bit of a thought experiment. **Harken back to the specific heat capacity questions you've done - what seems to stand out as 'missing' in each question?** The answer lies in how the physical attributes of the sample in question change - namely, the pressure and volume. When was the last time you ever considered it? Whenever the temperature of anything changes, will its pressure and volume change? Well, let's start first by treating the volume of the sample as constant. In reality, it's a mixture of both, but if we were to consider both dimensions we'd quickly go insane. Note the name of the **molar specific heat**. This implies to us that instead of being inversely proportional to the mass, the molar heat capacity will be inversely proportional to the number of moles in the system. So let's rewrite our equation: $Q = nC_{V}d T$ Where: - $Q$ **is the heat energy gained or lost by the sample** - $n$ **is the number of moles of particles in the sample** - $C_V$ **is the molar specific heat at constant volume** - $dT$ **is the instantaneous change in temperature.** Since we're now considering the number of particles in our expression, we can also find an expression for the internal energy in terms of $C_V$ by [[The First Law of Thermodynamics|the first Law of Thermodynamics]]: First, let us write the First Law and substitute the $Q$ term for the above expression: $dU = nC_{V}dT - W$ For this case, as $W = pdV$ and the volume does not change, no work is being done on the system and $W$ is 0. Therefore, we can write $C_V$ as: $C_{V} = \frac{dU}{ndT}$ Remember that $dU$ is also equal to $\frac{3}{2}nRT$; simplifying gives us our penultimate expression for $C_V$: $C_{V} = \frac{3}{2}R$ Therefore, if we substitute this value for $C$ back into the equation for internal energy $U$, we get: $U = nC_v dT$ ## The Isobaric Case Now imagine a case where the **pressure** stays the same. Now, work is being done on the system. The equation for the molar heat energy follows the same pattern as the two other equations preceding it: $Q = nC_{P}dT$ Where $C_P$ is the molar specific heat at a constant pressure. Since we have an equation for the internal in terms of $C_V$, let's use that to get an equation for $U$, and by extension $C_P$: $nC_{V}dT = nC_{P}dT - pdV$ Let's make a substitution of $pdV = nRdT$, as per the [[Kinetic Theory & Ideal Gases (Thermodynamics)#The Ideal Gas Law|Ideal Gas Law]]: $nC_{V}dT = nC_{P}dT - nRdT$ Now finding an expression for $C_P$ is as easy as cancelling out terms: $C_{V} = C_{P} - R$ Or: $R = C_{P} - C_{V}$ Done! %%problems tomorrow!%% $$ $$