Thermo - Heat & Calorimetry (Lesson)

Heat & Calorimetry

Introduction

The specific heat capacity quantifies the amount of heat required to raise the temperature of a substance by one degree Celsius. This helps chemists to understand how much the temperature of an object will increase or decrease by the gain or loss of heat energy. Materials that are good conductors of heat energy (like most metals) have a low specific heat and materials that are poor conductors of heat energy (insulators, like rubber) have a high specific heat. You will investigate this concept as well as chemical reactions that give off and take in heat energy in this lesson.

Thermal Conductivity and Specific Heat Capacity

In segment C, students learn about thermal conductivity and specific heat capacity and see if their predictions for how fast the ice cubes melt were correct. Our host explains how heat capacity affects the earth's climate, and the students begin an experiment using samples of greenhouse gases.

Download the note taking guide for Chemistry Matters Unit 8 Segment C. Links to an external site.

Download the key to the questions to consider for Chemistry Matters Unit 8 Segment C. Links to an external site.

Note: You will not be required to complete the greenhouse gas exploration, but pay close attention to the concepts explored by the students in the video.

Calculating Energy Changes - Heating and Cooling

Equations to Calculate Heat

-q_lost = q_gained

The negative sign on the q_lost shows that the heat is released from that object to its surroundings. This corresponds to a decrease in temperature and a negative ∆T. The positive q value for heat gained indicates that an object gained heat energy from its surroundings. This corresponds to an increase in temperature and a positive ∆T.

Heat Capacity

In addition to using -q _lost = q _gained, there are several other equations we will use to calculate heat. One of these equations uses heat capacity. Heat capacity is the amount of heat required to change the temperature of a substance by one degree. Heat capacity depends upon the size of the sample (therefore an extensive property) and the identity of the substance.

A chemical system is put in thermal contact with a heat bath. The heat bath is a substance, such as water, whose heat capacity has been well established by previous experiments. A process is initiated in the chemical system (heating/cooling, phase transition, or chemical reaction), and the change in temperature of the heat bath is measured.

Because the heat capacity of the heat bath is known, the observed change in temperature can be used to determine the amount of energy exchanged between the system and the heat bath.

The energy exchanged between the system and the heat bath is equal in magnitude to the change in energy of the system. If the heat bath increased in temperature, its energy increased, and the energy of the system decreased by this amount. If the heat bath decreased in temperature, and therefore energy, the energy of the system increased by this amount.

Make sure to visualize this setup as you read and solve calorimetry problems. It will help you solve the problem correctly and understand what is happening conceptually.

First organize your data.

***Note: The final temperature is the same for the metal and the water.**

Next, think about your master equation.

-Q_{lost by metal} = Q_{gained by water}

If you plug in the equation, q=CΔT

-C_(metal)ΔT_(metal) = C_(water)ΔT_(water)

Plug in what you know:

-C_(metal) (-190^oC) = C_(water)(5.0^oC)

In order to calculate C_(metal), we need to know C_(water). To do this, use the mass of the water and the definition of the calorie. The calorie is the amount of energy required to heat 1 gram of water 1 degree Celsius. Since we have 250g of water, 250 calories are required for each degree. We can make this into a conversion factor: 250cal = 1^oC. Heat capacity has the unit J/^oC. So, we will also need to convert the energy to J, remembering the conversion factor: 1cal = 4.184J.

C(water) = (250cal/1^oC) (4.184J/1cal) = 1046J

Now we can plug this into the master equation above

-C_(metal)(-190^oC) = (1046J)(5.0^oC)

-C_(metal) = 27.5 J/^oC

Specific Heat Capacity

Specific heat capacity is an intensive property that is inherent to that substance, and does not change with the size of the sample. It is defined as the amount of energy required to raise the temperature of 1 gram of a substance 1 degree Celsius. It is important to understand that the heat capacity of a drop of water is much lower than the heat capacity of a pool of water, while their specific heats are the same. Note in the table here of some specific heats that the specific heat of water is much larger than other substances. You will be expected to memorize the specific heat of water. The value typically used is 4.184 J/g°C.

Specific Heat Problems

Example 1: How much heat is required to raise the temperature of 53g of water from 11^oC to 44^oC?

• To solve for heat when the temperature changes, use the equation: Q = m x C x ∆T
• In this problem, the mass is given, m = 53g
• If the substance is known, the value of C can be found on a chart like the one above.
  • C = 4.184 J/g ^oC
• The change in temperature is found by subtracting the initial temperature from the final temperature.

ΔT = T_f – T_i = 44^oC – 11^oC = 33^oC

To calculate the amount of heat (Q), plug the above values into the equation

• Q = m x C x ΔT
• Q = (53 g) x (4.184 J/g^oC) x (33^oC) = 7300J

Example 2: How much heat is released when 21 g of Al cools from 31.0 ^oC to 27.0 ^oC?

Solution: Q = m x C x Δt = (21 g) x (0.89 J/g^oC) x (27.0 oC – 31.0 ^oC) =-75J

Note: Did you get a negative number for ΔT? If so, you did it right! Δt = 27^oC -31^oC = -4.0^oC. This negative number just means that heat was released.

Example 3: What is the specific heat capacity of 32.3 g of a metal if 50.0 J of heat increases the temperature of the metal by 3.5^oC?
Solution:

• Q = m x C x ΔT
• C = Q / (m ΔT)
• C = 50.0J / {(32.3g) (3.5 ^oC)} = 0.44 J/g^oC

Example 4: What mass of water releases 2,620 J of heat when cooled 12.5^oC.

Solution:

• Q = m x C x ΔT
• m = Q/(C Δt)
• m = 2,620J/{(4.184J/goC)(12.5^oC)} = 50.1g

Example 5: Three liquids of the same mass absorb the same amount of heat. Liquid A’s temperature rises 20^oC, liquid B’s rises 10^oC, and liquid C’s doesn’t change. Explain.

Solution: Liquid A has the lowest specific heat capacity of all three and therefore does not require as much heat to raise its temperature. B has the next highest and C has the highest of all three – so high in fact that the temperature does not increase at all. It is also possible that liquid C is changing phase (boiling).

Specific Heat Capacity Practice

Calculating Energy Change - Phase Change

Hopefully you will remember from previous units that when heat energy is supplied to a solid, the temperature of the solid climbs steadily until the melting point is reached. Once the melting point is reached, even though we are still supplying heat energy to the system, the temperature remains constant while the phase changes from solid to liquid. After the phase change is complete, the temperature starts to climb again. This is represented on a heating curve like the one shown here.

Energy must be transferred to a system to cause it to melt or boil. The energy of the system therefore increases as the system undergoes a solid-liquid (or liquid-gas) phase change. Likewise, a system gives off energy when it freezes (or condenses). The energy of the system decreases as the system undergoes a liquid-solid (or gas-liquid) phase change.

The heat energy which a solid absorbs when it melts is called heat of fusion. (The word fusion means the same thing as melting. This sounds backwards, doesn't it? Just think of fusion as the word to describe the liquid/solid phase change. It might be easier to remember fusion if you think about ice forming from liquid water.) Heat of fusion is also called enthalpy of fusion. The term enthalpy will be defined on another page. Heat of fusion has the symbol ΔH fus. (H is the symbol for enthalpy.) ΔH fus is most often expressed in kJ or in kJ/mol. When it is expressed in kJ/mol, it is referred to as molar heat of fusion.

The amount of energy released when the same liquid changes into a solid is the same magnitude as ΔH fus, but it has the opposite sign, indicating that energy was released.

The amount of energy needed to change a liquid into a gas (boiling) is called heat of vaporization. The amount of energy needed to vaporize one mole of a pure substance is called molar heat of vaporization, and the energy released in condensation has an equal magnitude.

Relating Heat of Vaporization to Intermolecular Attraction (IMA)

Heat energy is absorbed when a liquid boils because molecules which are held together by mutual attraction in the liquid are separated from each other as the gas is formed. Such a separation requires energy. The energy needed to separate the molecules in a liquid depends on the magnitude of the intermolecular forces (IMA).

Notice in the chart that the heats of vaporization of water and ammonia are very large, which is just what we would expect for hydrogen-bonded substances. By comparison, CH4, a nonpolar substance composed of atoms of similar size, has a very small heat of vaporization. Note also that polar substances such as HCl and SO2 have fairly large heats of vaporization compared to nonpolar substances. For example, compare HCl to Cl2. Even though Cl2 contains two relatively large atoms, and therefore would be expected to have larger London forces than HCl, the latter substance has the larger ∆Hvap. This must be due to dipole-dipole attractions between polar HCl molecules—attractions that are absent in nonpolar Cl2. We can thus expect liquids with strong intermolecular forces to have larger heats of vaporization.

Similar trends can be seen in heats of fusion as well. Solids, like ice, that have strong intermolecular force s have much higher molar heats of fusion than those like CH4 with weak ones.

Intermolecular Attractions Practice

Calculations with Heat of Fusion and Heat of Vaporization

Equations to Calculate Heat: -q_lost = q_gained

The negative sign on the qlost shows that the heat is released from that object to its surroundings. This corresponds to a decrease in temperature and a negative ∆T. The positive q value for heat gained indicates that an object gained heat energy from its surroundings. This corresponds to an increase in temperature and a positive ∆T.

The specific heat of fusion represents the amount of heat needed to change the state of a substance from solid to liquid on a per amount basis. It is reported in J/g. Likewise, the specific heat of vaporization is reported in J/g. These values are useful when we are given a specific mass of a substance and are asked to calculate how much energy is absorbed or released during a phase change. Some common values are listed in the table here.

Calculations with Heat of Fusion and Heat of Vaporization Practice

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