Heat and Temperature

10.2.2 Temperature

What heat means in thermodynamics, and how we can calculate heat using the heat capacity.

We can calculate the heat released or absorbed using the specific heat capacity C, the mass of the substance m, and the change in temperature ΔT in the equation:

q = m × C × ΔT

https://www.khanacademy.org/science/chemistry/thermodynamics-chemistry/internal-energy-sal/a/heat

Heat in thermodynamics
What contains more heat, a cup of coffee or a glass of iced tea? In chemistry class, that would be a trick question (sorry!). In thermodynamics, heat has a very specific meaning that is different from how we might use the word in everyday speech. Scientists define heat as thermal energy transferred between two systems at different temperatures that come in contact. Heat is written with the symbol q or Q, and it has units of Joules (J\text JJJ).
Three melting ice cubes in a puddle of water on a mirrored surface.
As ice melts, heat is transferred from the water to the surroundings. Photo of ice cubes from flickr, CC BY 2.0.
Heat is sometimes called a process quantity, because it is defined in the context of a process by which energy can be transferred. We don’t talk about a cup of coffee containing heat, but we can talk about the heat transferred from the cup of hot coffee to your hand. Heat is also an extensive property, so the change in temperature resulting from heat transferred to a system depends on how many molecules are in the system.
Relationship between heat and temperature
Heat and temperature are two different but closely related concepts. Note that they have different units: temperature typically has units of degrees Celsius (∘C^\circ\text C​∘​​Cdegree, C) or Kelvin (K\text KKK), and heat has units of energy, Joules (J\text JJJ). Temperature is a measure of the average kinetic energy of the atoms or molecules in the system. The water molecules in a cup of hot coffee have a higher average kinetic energy than the water molecules in a cup of iced tea, which also means they are moving at a higher velocity. Temperature is also an intensive property, which means that the temperature doesn’t change no matter how much of a substance you have (as long as it is all at the same temperature!). This is why chemists can use the melting point to help identify a pure substance−-−minusthe temperature at which it melts is a property of the substance with no dependence on the mass of a sample.
On an atomic level, the molecules in each object are constantly in motion and colliding with each other. Every time molecules collide, kinetic energy can be transferred. When the two systems are in contact, heat will be transferred through molecular collisions from the hotter system to the cooler system. The thermal energy will flow in that direction until the two objects are at the same temperature. When the two systems in contact are at the same temperature, we say they are in thermal equilibrium.
Zeroth law of thermodynamics: Defining thermal equilibrium
The zeroth law of thermodynamics defines thermal equilibrium within an isolated system. The zeroth law says when two objects at thermal equilibrium are in contact, there is no net heat transfer between the objects; therefore, they are the same temperature. Another way to state the zeroth law is to say that if two objects are both separately in thermal equilibrium with a third object, then they are in thermal equilibrium with each other.
The zeroth law allows us to measure the temperature of objects. Any time we use a thermometer, we are using the zeroth law of thermodynamics. Let’s say we are measuring the temperature of a water bath. In order to make sure the reading is accurate, we usually want to wait for the temperature reading to stay constant. We are waiting for the thermometer and the water to reach thermal equilibrium! At thermal equilibrium, the temperature of the thermometer bulb and the water bath will be the same, and there should be no net heat transfer from one object to the other (assuming no other loss of heat to the surroundings).
Heat capacity: Converting between heat and change in temperature
How can we measure heat? Here are some things we know about heat so far:

When a system absorbs or loses heat, the average kinetic energy of the molecules will change. Thus, heat transfer results in a change in the system’s temperature as long as the system is not undergoing a phase change.
The change in temperature resulting from heat transferred to or from a system depends on how many molecules are in the system.

We can use a thermometer to measure the change in a system’s temperature. How can we use the change in temperature to calculate the heat transferred?
In order to figure out how the heat transferred to a system will change the temperature of the system, we need to know at least 2222 things:

The number of molecules in the system
The heat capacity of the system

The heat capacity tells us how much energy is needed to change the temperature of a given substance assuming that no phase changes are occurring. There are two main ways that heat capacity is reported. The specific heat capacity (also called specific heat), represented by the symbol c\text ccc or C\text CCC, is how much energy is needed to increase the temperature of one gram of a substance by 1 ∘C1~^{\circ}\text C1 ​∘​​C1, space, degree, C or 1K1\,\text K1K1, space, K. Specific heat capacity usually has units of Jgrams⋅K\dfrac{\text J}{\text{grams}\cdot\text K}​grams⋅K​​J​​start fraction, J, divided by, g, r, a, m, s, dot, K, end fraction. The molar heat capacity, Cm\text C_\text mC​m​​C, start subscript, m, end subscript or Cmol\text C_{\text{mol}}C​mol​​C, start subscript, m, o, l, end subscript, measures the amount of thermal energy it takes to raise the temperature of one mole of a substance by 1 ∘C1~^{\circ}\text C1 ​∘​​C1, space, degree, C or 1K1\,\text K1K1, space, K, and it usually has units of Jmol⋅K\dfrac{\text J}{\text{mol}\cdot\text K}​mol⋅K​​J​​start fraction, J, divided by, m, o, l, dot, K, end fraction. For example, the heat capacity of lead might be given as the specific heat capacity, 0.129Jg⋅K0.129\,\dfrac{\text J}{\text{g}\cdot\text K}0.129​g⋅K​​J​​0, point, 129, space, start fraction, J, divided by, g, dot, K, end fraction, or the molar heat capacity, 26.65Jmol⋅K26.65\,\dfrac{\text J}{\text{mol}\cdot\text K}26.65​mol⋅K​​J​​26, point, 65, space, start fraction, J, divided by, m, o, l, dot, K, end fraction.
[Why do different substances have different specific heat capacities?]
Let’s think about what is happening on a molecular level when we add thermal energy to some molecules. The thermal energy can be stored as vibrations and rotations between atoms within a molecule, which does not significantly increase the temperature of the system. The energy can also be used to disrupt intermolecular interactions and increase the velocity of the entire molecule, which increases the translational kinetic energy of the molecule.
Temperature is primarily a measure of the translational kinetic energy of the system. Depending on the molecular structure and intermolecular interactions, different substances can store different amounts of thermal energy as vibrations and rotations before the temperature increases.
Calculating q\text qqq using the heat capacity
We can use the heat capacity to determine the heat released or absorbed by a material using the following formula:

q=m×C×ΔT\text q = \text {m} \times \text C \times \Delta \text Tq=m×C×ΔTq, equals, m, times, C, times, delta, T

where m\text{m}mm is the mass of the substance (in grams), C\text{C}CC is the specific heat capacity, and ΔT\Delta \text TΔTdelta, T is the change in temperature during the heat transfer. Note that both mass and specific heat capacity can only have positive values, so the sign of q\text qqq will depend on the sign of ΔT\Delta \text TΔTdelta, T. We can calculate ΔT\Delta \text TΔTdelta, T using the following equation:

ΔT=Tfinal−Tinitial\Delta \text T=\text T_{\text{final}}-\text T_{\text{initial}}ΔT=T​final​​−T​initial​​delta, T, equals, T, start subscript, f, i, n, a, l, end subscript, minus, T, start subscript, i, n, i, t, i, a, l, end subscript

where Tfinal\text T_{\text{final}}T​final​​T, start subscript, f, i, n, a, l, end subscript and Tinitial\text T_{\text{initial}}T​initial​​T, start subscript, i, n, i, t, i, a, l, end subscript can have units of either ∘C~^{\circ}\text C ​∘​​Cspace, degree, C or K\text KKK. Based on this equation, if q\text qqq is positive (energy of the system increases), then our system increases in temperature and Tfinal>Tinitial\text T_{\text{final}}>\text T_{\text{initial}}T​final​​>T​initial​​T, start subscript, f, i, n, a, l, end subscript, is greater than, T, start subscript, i, n, i, t, i, a, l, end subscript. If q\text qqq is negative (energy of the system decreases), then our system’s temperature decreases and Tfinal<tinitial\text t_{\text{final}}<\text=”” t_{\text{initial}}t​final​​<t​initial​​t,=”” start=”” subscript,=”” f,=”” i,=”” n,=”” a,=”” l,=”” end=”” is=”” less=”” than,=”” t,=”” subscript.<br=””>Example problem: Cooling a cup of tea
Let’s say that we have 250mL250\,\text{mL}250mL250, space, m, L of hot tea which we would like to cool down before we try to drink it. The tea is currently at 370K370\,\text K370K370, space, K, and we’d like to cool it down to 350K350\,\text K350K350, space, K. How much thermal energy has to be transferred from the tea to the surroundings to cool the tea?
A cup of black tea with a slice of lemon in a white teacup with a saucer.
The hot tea will transfer heat to the surroundings as it cools. Photo from Photozou, CC BY-NC-ND 2.5
We are going to assume that the tea is mostly water, so we can use the density and heat capacity of water in our calculations. The specific heat capacity of water is 4.18Jg⋅K4.18\,\dfrac{\text J}{\text g \cdot \text K}4.18​g⋅K​​J​​4, point, 18, space, start fraction, J, divided by, g, dot, K, end fraction, and the density of water is 1.00gmL1.00\,\dfrac{\text g}{\text {mL}}1.00​mL​​g​​1, point, 00, space, start fraction, g, divided by, m, L, end fraction. We can calculate the energy transferred in the process of cooling the tea using the following steps:
1. Calculate the mass of the substance
We can calculate the mass of the tea/water using the volume and density of water:

m=250mL×1.00gmL=250g

2. Calculate the change in temperature, ΔT\Delta \text TΔTdelta, T
We can calculate the change in temperature, ΔT\Delta \text TΔTdelta, T, from the initial and final temperatures:

ΔT=Tfinal−Tinitial=350K−370K=−20K\begin{aligned}\Delta \text T&=\text T_{\text{final}}-\text T_{\text{initial}}\\ \\ &=350\,\text K-370\,\text K\\ \\ &=-20\,\text K\end{aligned}​ΔT​​​​​​​=T​final​​−T​initial​​​=350K−370K​=−20K​​

Since the temperature of the tea is decreasing and ΔT\Delta \text TΔTdelta, T is negative, we would expect q\text qqq to also be negative since our system is losing thermal energy.
3. Solve for q\text qqq
Now we can solve for the heat transferred from the hot tea using the equation for heat:

q=m×C×ΔT=250g×4.18Jg⋅K×−20K=−21000J

Thus, we calculated that the tea will transfer 21000J21000\,\text J21000J21000, space, J of energy to the surroundings when it cools down from 370K370\,\text K370K370, space, K to 350K350\,\text K350K350, space, K.
Conclusions
In thermodynamics, heat and temperature are closely related concepts with precise definitions.

Heat, q\text qqq, is thermal energy transferred from a hotter system to a cooler system that are in contact.
Temperature is a measure of the average kinetic energy of the atoms or molecules in the system.
The zeroth law of thermodynamics says that no heat is transferred between two objects in thermal equilibrium; therefore, they are the same temperature.
We can calculate the heat released or absorbed using the specific heat capacity C\text CCC, the mass of the substance m\text mmm, and the change in temperature ΔT\Delta \text TΔTdelta, T in the following equation:

q=m×C×ΔT\text q = \text {m} \times \text C \times \Delta \text Tq=m×C×ΔT