Temperature and Heat Study Guide

Example

Find the conversion from Fahrenheit to Kelvin scale. Find the value of 32° F in Kelvin degrees and check your work.

Solution

We have already determined that the conversIOn between Fahrenheit and Celsius is:

We can use this equation to solve for t(°C) as a function of t(°F) and then use the result in T(K).

With this expression, we go back to T(K) as a function of t(°C).

t(°F) = 32°F

T(K) = 273.15K

or

t(°C) = 0°C

The result agrees with our initial discussion about the Kelvin scale and its direct connection with the Celsius scale.

Heat

For a system to change its temperature, the object exchanges energy with its surroundings. We call this energy heat, and we measure heat in joules (from James Joule, 1818–1889).

We have seen previously that temperature is specific to a certain state. Therefore, we call temperature a quantity of state. In contrast, heat is an energy flow established when there is a thermal contact between two different temperature states. We call heat a quantity of process.

Heat is related to a measure of the change in the motion and interaction of the particles, which is called internal energy. Internal energy is equal to the kinetic energy and the potential energy of the particles. Kinetic energy measures the translational, rotational, and vibrational motion of the atoms and molecules in the object, whereas potential energy measures the interaction between the particles. When the temperature increases, the kinetic energy increases. One way to accomplish this process is by exchanging heat with the object.

Exchanging Heat

Heat will flow freely from a high-temperature object to a low-temperature object because of the difference in temperature.

Heat Transfer

We have seen why heat flows; let's learn now about the means of this flow. On an atomic scale, materials in different states are built differently: Solids have an internal structure, and the atoms and molecules are bound through strong bonds. Disturbing the bonds in one place will create a disturbance in the lattice, and so we end up with a propagation of the initial effect. Liquids and gases are different in the sense that the structure is not isotropic (the same in all directions, or completely absent as in the case of gases). Hence, a disturbance in one side of the container with fluid will spread differently than it does in a solid.

There are three types of heat transfer: conduction, convection, and radiation.

In the case of conduction, the heat is transferred through the material itself, and a difference in temperature between different sides of the object is required. On an atomic scale. the particles on the side of the object where the temperature is larger are characterized by a greater kinetic energy. While moving, they will collide with slower particles and, in the process, lose some of their kinetic energy to the slower moving particles that now accelerate.

Metals are good thermal conductors because of the free electrons that move through the lattice. Other materials such as glass, plastic, and paper are poor conductors due to the light interaction between constituent particles. Still other materials, such as gases, are isolators due to the large distance between particles.

The situation is different in convection, where the transfer of heat happens due to movement of the substance through space. Consider forcing warm air into a room through floor-level inlets. What happens in time? The warm air rises, and the cold air sinks. Warm air has atoms and molecules that move faster, and they are farther apart; therefore, the density is less than the density of cold air. And, as we have seen in the last lesson, a lower density material will be buoyed up.

A different process that does not require contact is called radiation. With this transfer, heating is accomplished by electromagnetic radiation. Every object with a temperature more than zero absolute Kelvin radiates infrared radiation, which in turn is absorbed by other objects and increases their temperature.

Heat and Temperature Change

An object that is warmed up will experience an increase in temperature. The variation of temperature is different depending on the nature of the object. To characterize this dependence, we define two coefficients: the heat capacity and the specific heat. The coefficients of heat capacity and specific heat for different materials are tabulated.

If a quantity of heat Q is transferred to a substance thereby increasing its temperature by ΔT = T₂–T₁ the heat capacity is defined to be:

This coefficient is not dependent on the mass of the specific object and is measured in J/K or J/C°

If a quantity of heat Q is transferred to a substance of mass m and is increasing its temperature by ΔT = T₂ – T₁ , the specific heat is defined to be:

We can also rewrite the equation as:

Q = m · c · (T₂ – T₁)

Using this equation, we define a new unit for heat called the calorie. One calorie (1 cal) is the heat necessary to be transferred to 1 g of water to increase its temperature by one Celsius degree (from 14.5 to 15.5°C). The conversion from calories to joules is:

1 cal = 4.186 J

The nutritional calorie that you see on food labels is actually 1,000 calories and is symbolized by C.

1 C = 1,000 cal

Another usual unit is one British thermal unit (BTU), which is the heat necessary to be transferred to 1 pound of water to raise its temperature from 63 to 64° F.

Some containers are built such that they make good isolators. The heat transferred to a fluid is completely exchanged with the fluid and none lost to its surroundings. Such containers are called calorimeters, and the study of the heat exchange in these systems is called calorimetry. In this case, the heat coming from a hot reservoir is completely transferred to the cold reservoir:

Q_hot = –Q_cold

The minus sign indicates that the system that cools down looses energy (heat is coming out), and therefore, the heat transfer is negative. This is called the calorimetric equation.

Some of the most usual materials encountered and their specific heats are shown in Table 11.1.

Example

An aluminum piece of 400 g is placed in a container that holds 100 g water at 80° C. The water cools down to 20° C. In the process, the aluminum piece gets warmer and reaches a temperature of 45° C. What was the initial temperature of the aluminum piece? Consider the only heat exchange to be between the aluminum and the water.

Solution

First, convert all quantities to 51 units. Next, set the equations and solve for the initial temperature.

m_al = 400 g = 400 g ·1 kg/1,000 g = 0.4 kg

m_water = 100 g = 100 g · 1 kg/1,000 g = 0.1 kg

t_{water initial} = 80° C

t_{water final} = 20° C

t_{Al final} = 45° C

t_{al initial} = ?

Because the system is thermally isolated, the heat released by the water is absorbed by the aluminum.

Q_water = – Q_aluminum

Q_water = m_water· C_water · (t_{water final} – t_{water initial})

Q_water = 0.1 kg · 4,186 J/kg · °C · (80° C – 20° C)

Q_water = 0.1· 4,186 J · 60

Q_water = 2.5 . 10⁴ J

–Q_aluminum = –m_Al · c_Al · (t_{Al final} – t_{Al initial})

–Q_aluminum == – 0.4 kg . 900 j/kg °C · (45 – t_{Al initial})

2.5 · 10⁴ J = 0.4 · 900 J/ °C · (45 – t_{Al initial})

2.5 · 10⁴ J/(0.4 · 900 J/°C) = (45 – t_{Al initial})

2.5 · 10⁴°C/(0.4 · 900) = (45 – t_{Al initial})

70 °C = (45 – t_{Al initial})

t_{Al initial} = (45 – 70°)C = – 25°C

t_{Al initial} = – 25°C

Heat and Phase Change

Is the heat absorbed or released by an object always changing its temperature? The answer is no. Think, for example, about boiling water: Once the water boils, there is another phenomenon taking place called vaporization. So, in this example, the heat absorbed is used first to increase the water temperature and then to change the phase from liquid to gas. This transformation is called a phase change, and in our example, the pressure is considered to be constant because the transformation takes place in open space. The heat exchanged when the liquid is experiencing a phase change is called latent heat. Measurements show that in a phase change, the temperature is constant until the transformation to another phase happens in the bulk of the substance.

There are a few processes regarding phase changes: from solid to liquid and the inverse. These are called melting and freezing. Together, they are called fusion; specifically, they are vaporization at evaporation or condensation, and sublimation at the phase change from solid to gas or gas to solid.

The heat exchanged in a phase change has the same value regardless of the direction of change: The coefficient of latent heat of melting is equal to the coefficient of latent heat of freezing.

In many books, these processes are summarized in a graph (see Figure 11.2) of the temperature dependence of the absorbed energy (heat).

If you interpret the graph, you will see that the slopes of the graphs for ice and for steam are almost the same. All things being equal, this translates to the ratio of heat and temperature change, which gives us the specific heat. If you check with the table, you will see the two coefficients close in value. How about the water phase? The slope is smoother in this case, as shown also by the numbers in the coefficients of specific heat.

In a calorimetry measurement, the hot reservoir or/and the cold reservoir heat might contain terms similar to m · c · ΔT as we have already seen before. However, there are new terms involving a process with no temperature change, a process called phase transition. Let's try to define a measure of the transition from the point of view of heat transfer.

Coefficient of Latent Heat

If a quantity of heat Q is transferred to a substance of mass m and the substance has a phase change, then the coefficient of latent heat L is given as follows:.

This coefficient depends on the nature of the substance and on the type of phase change occurring.

Example

An aluminum piece of 400 g is placed in a container that holds 100 g of ice at –10°C . The water warms up to 5°C. In the process, the aluminum piece gets colder and reaches a temperature of 45°C. What was the initial temperature of the aluminum piece? The experiment takes place in a calorimeter.

Solution

First, convert all quantities to SI units. Next, set the equations and solve for the initial temperature.

According to the problem, the exchange happens in a calorimeter. Therefore, the only heat exchange happens between the aluminum and the ice.

m_al = 400 g = 400 g ·1 kg/1,000 g = 0.4 kg

m_ice = 100 g = 100 g ·1 kg/l,000 g = 0.1 kg

t_{ice initial} = – 10° C

t_{water final} = 5°C

t_{Al final} = 45° C

t_{al initial} = ?

Because the system is thermally isolated, the heat released by the aluminum is absorbed by the ice and then the water.

Q_ice = –Q_aluminum

The process of heating of water goes through three different steps: First, the ice absorbs heat and the temperature increases from –10 to 0° C; second, the ice melts and there is no change in temperature; and last, the water warms up from 0 to 5° C. So we will have three terms referring to the ice to water heat absorption.

Q_ice = Q_{– 10 to 0° C} + Latent heat of melting + Q_{0 to 5° C}

Q_ice = m_ice · C_ice ·(0°C – t_{ice initial}) + m · L + m_ice · C_water · (t_{water final} – 0° C)

(L = 3.33 × 105 J/kg)

Q_ice = m_ice · C_ice · (0 – ( –10)) + m_ice · L + m_ice · C_water · (5 – 0)

Q_ice = 3.74 . 10⁴J

Q_aluminum = m_a1 · c_al · (t_{al final} – t_{al intial})

Q_aluminum = 0.4 kg . 900 J/kg · °C· (45 – t_{Al initial})

3.74 · 10⁴J = – 0.4 · 900/J°C · (45 – t_{Al initial})

3.74 · 10⁴ J/(0.4 · 900J/°C) = – (45 – t_{Al initial})

3.74 · 10⁴⁰ C/(0.4 · 900) = –(45 – t_{Al initial})

104°C = (45 – t_{Al initial})

t_{Al initial} = (45 +104)°C = 149°

t_{Al initial} = 149°C

Thermal Expansion

We have now discussed the effect of heat exchange on a solid, and we learned that in the process, the particles inside a solid will move apart from each other as they acquire kinetic energy (due to an increase of average speed). The effect is easily observed if you make a careful measurement of the length of a rod that is subjected to increased temperature. This process of expansion is dependent not only on the amount of energy exchanged but also on the nature of the material.

In order to have a quantitative explanation of how substances expand, a few characteristics should be defined and measured: ΔL = L – L₀ = change in length; α = coefficient of thermal expansion (1/°C); L₀ = initial length; ΔT = change in temperature.

The expansion along one direction is called linear thermal expansion and can be related to these coefficients and parameters through the following expression:

ΔL = α · L₀ · ΔT

But the structure of a solid is three dimensional so we cannot assume that the process of thermal expansion happens linearly. All x, y, and z directions will be affected by the expansion. A simple way to correlate 3–D expansion with linear expansion is to imagine a cube of initial side L₀ that is heated up to a temperature T_o + ΔT. All the sides of the cube will expand, and the change in volume be found by subtracting the initial and final volumes. The volume thermal expansion then is characterized by the following equation:

ΔV = β · V₀ Δ T

where ΔV = change in volume; β = coefficient of volume thermal expansion; V₀ = initial volume; ΔT = change in temperature. If we consider that:

ΔV = V – V₀ = L³ – L₀³

and we write an expression for L as a function of L₀ as given by

ΔL = L – L₀ = α · L₀ · Δ T

then we can show that by neglecting higher order terms in α (measurement shows this coefficient being very small; therefore, the higher orders will be even smaller. They can be neglected):

β = 3 · α

And:

ΔV = 3 · α · V_o · ΔT

Example

A copper cube of size 25 cm has a coefficient of linear expansion of 17 · 10 ^{– 6°} C^–1 Find the change in length and volume if the temperature changes 12 °C from morning to noon.

Solution

First, convert the units to SI. Next, build our equations to determine the unknown.

α = 17 · 10 ^{– 6°} C^–1

ΔT = 12°C

L₀ = 25 cm = 0.25 m

ΔL = ?

ΔV = ?

We can determine first the change in length:

ΔL = α · L₀ · Δ T =

17 · 10 ^{– 6°} C^–1 · 0.25 m · 12°C = 5.1 · 10^–5m

ΔL = 5.1 · 10^–5m

ΔV = 3 · α · V₀ · Δ T

V₀ = L₀3

ΔV = 3 · α · L₀3 · Δ T

3 . 17 · 10^{– 6°} c^{– 1} · (0.25 m)³ ·12°C

9.6 · 10 ^{– 6}m³

ΔV = 9.6 · 10 ^{– 6}m³

You can see that the change is small in both cases but is definitely something measurable.

Practice problems of this concept can be found at: Temperature and Heat Practice Questions