HS Chemistry - Heat in Reactions

Calorimetry

Overview of The Page

This page will cover:

• What is specific heat capacity? What is its related equation?
• What is specific latenet heat? What is its related equation?

A calorimeter can be used to calculate:

• Specific heat capacity of a substance, which is how much energy must be provided to the substance in order to raise the temperature of 1 gram/kilogram of that substance by 1°C

• Specific latent heat of a substance, which is how much energy must be provided to the substance in order to change the state of matter of 1 gram/kilogram of that substance without increasing its temperature.

Equations for heat transfer

For the most part, as you heat something up, the particles gain more energy and move faster in their phase of matter $solid, liquid, gas, plasma$. That is, if you heat a solid, the particles will vibrate faster, but the substance will remain a solid, so the particles won't suddenly start moving around with all this newfound energy.

The equation for that is:

Q=mC∆T

Variable	What It Stands For
Q	The thermal energy provided to the substance, in Joules or kiloJoules $depends on what you're given, and what unit C is$
m	The mass of the substance in grams or kilograms $depends on what you're given, and what unit C is$
C	The specific heat capacity of the substance. It uses one of the following units: J/g°C J/kg°C $1 J/kg°C is equal to 1/1000 J/g°C$ kJ/g°C $1 kJ/g°C is equal to 1000 J/g°C$ kJ/kg°C $1 kJ/kg°C is equal to 1 J/g°C, so this unit is not commonly used$
∆T	The change in temperature of the substance: It is calculated by subtracting the initial temperature of the substance from the final temperature of the substance $∆T~f~ − ∆T~i~$ A positive ∆T value indicates that the temperature of the substance increased, and a negative ∆T value indicates that the temperature of the substance decreased

Q=mC∆T is the equation used when thermal energy is added to a substance that remains in a certain state of matter, but if heat keeps getting added or removed from it, eventually it is expected to change phases. At that point, when it is changing phases, thermal energy keeps getting added to it, but there's no increase in temperature because all the additional thermal energy is going into breaking the intermolecular forces $which is the primary determiner of what phase of matter something is in$.

The above equation doesn't work in that case, because although thermal energy is being put in, ∆T is 0, and if we used the equation with that value, we would get Q=0, which isn't right at all, because thermal energy is being added.

Therefore, we need another equation for when a phase change is occurring.

Q=mL

Variable	What It Stands For
Q	The thermal energy provided to the substance, in Joules or kiloJoules $depends on what you're given, and what unit L is$
m	The mass of the substance in grams or kilograms $depends on what you're given, and what unit L is$
L	The specific latent heat of the substance. It uses one of the following units: J/g J/kg $1 J/kg is equal to 1/1000 J/g°C$ kJ/g $1 kJ/g is equal to 1000 J/g$ kJ/kg $1 kJ/kg is equal to 1 J/g, so this unit is not commonly used$

The thermal energy supplied to the substance during a phase change is called latent heat. If it is changing from a solid to a liquid $or vice versa$, it is called the latent heat of fusion. If it is changing from a liquid to a gas $or vice versa$, it is called the latent heat of vaporization. If it is changing from a solid to a gas $or vice versa$, it is called the latent heat of sublimation.

Although different names may sometimes be used for the latent heat from a liquid to a gas and the latent heat from a gas to a liquid, their values are the same, except that if the substance is going from a higher-energy state to a lower-energy state, the specific latent heat $L$ has a negative sign rather than a positive sign. This causes Q to be negative as well, showing the direction of the transfer of heat.

If the substance is going from a lower-energy state to a higher-energy state, the specific latent heat $L$ has a positive sign. This causes Q to have a positive value as well, showing the direction of the transfer of heat.

It is important to remember that the mass of a substance is never negative.

How a calorimeter uses these equations

When two objects are in thermal equilibrium, they have the same temperature, and there is no net transfer of heat between them.

According to the first law of thermodynamics, energy cannot be converted or destroyed, only converted from one form to another.

A calorimeter uses these two principles to allow the user to calculate the specific heat capacity or specific latent heat of a substance.

The calorimeter insulates its contents from the outside, so that the amount of heat inside the calorimeter remains constant (to an experimentally allowed margin of error). It then involves placing a heated substance whose specific heat capacity/specific latent heat is unknown inside or next to one that is known. The temperatures and masses of both substances are taken at the beginning of the experiment, before placing them both inside the calorimeter. After a while, both substances achieve thermal equilibrium, and the temperatures of both substances are taken again.

Since energy cannot be created or destroyed, the amount of heat inside the calorimeter at the beginning of the experiment is the same as the amount of heat inside the calorimeter at the end of the experiment.

Since they're at thermal equilibrium now, they have the same amount of thermal energy.

Therefore:

Q_s1 = Q_s2

Where s1 is one substance inside the calorimeter $which we are measuring$ and s2 is the other substance inside the calorimeter $which we are measuring$.

1) If the experiment doesn't include phase changes $like placing a substance in water$, then it can be expanded to:

m_s1 × C_s1 × ∆T_s1 = m_s2 × C_s2 × ∆T_s2

Which further expands out to:

m_s1 × C_s1 × (T_s1f − T_s1i) = m_s2 × C_s2 × (T_s2f − T_s2i)

At this point, we know the masses of both substances and the initial and final temperatures of both substances, having measured them. The only other variables in the equation are C_s1 and C_s2. If we know one of them, we can calculate the other.

2) If one of the substances changes phases once in the experiment, then the equation can be expanded to:

m_s1 × C_s1 × ∆T_s1 = m_s2 × C_s2 × ∆T_s2 + m_s2 × L_s2

Which further expands out to:

m_s1 × C_s1 × (T_s1f − T_s1i) = m_s2 × C_s2 × (T_s2f − T_s2i ) + m_s2 × L_s2

At this point, we know the masses of both substances and the initial and final temperatures of both substances, having measured them. The only other variables in the equation are C_s1, C_s2, and L_s2. If we know two of them, we can calculate the third.

This can be expanded out for more phase changes as well.