# Physics Tutorial: Absorption of Heat

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In this Physics tutorial, you will learn:

• What are good and bad absorbers of heat?
• How does the heat travel between objects?
• What are the factors affecting the amount of heat absorbed by a substance?
• What is the specific heat capacity?
• What is the conversion factor between units of heat energy?
• What are the main features of the stable states of matter?
• What are the phases of state change?
• What is latent heat and why it is called so?
• What is the specific latent heat of fusion and vaporization?
• How to calculate the amount of heat absorbed or released by a substance?

## Introduction

What happens to the temperature of an object when it is heated? Do you think different materials need the same amount of heat energy to increase their temperature at the same rate? Why?

How many states of matter do you know? How do they differ from each other?

What happens to the molecular bonds during the change in state of matter?

In this tutorial, we will discuss about everything related to heat energy and its effect on the structure of matter. This is a tutorial in which the behaviour of matter when it absorbs or releases heat energy will be explained in molecular terms.

## Absorption of Heat. Good and Bad Absorbers of Heat.

Heat is a kind of energy, as explained in the first tutorial of this section "Temperature. The Zeroth Law of Thermodynamics". It is an energy of transfer, i.e. it appears only when thermal energy is transferred from one object or system into another. We can identify which of two objects in contact absorbs heat by measuring their temperature. The hottest object (the one with the highest temperature) releases heat, which is absorbed by the coldest object. This is an universal law of physics, i.e.

"Heat energy always flows from the hottest object to the coldest one."

However, the absorption of heat energy is not equal for all materials. Some materials can absorb the heat more easily (good absorbers of heat) while some other materials show resistance to heat absorption (bad absorbers of heat). As a result, if we provide the same amount of energy to each of these materials, the temperature of good absorbers increases more than that of bad absorbers for the same amount of matter.

The factors affecting the amount of heat absorbed by a substance are:

1. The amount of substance. Mathematically, this factor is represented through the mass of substance, m.
2. The increase in temperature of the substance. This factor is represented in formulae by the symbol ΔT = T - T0, where T is the final temperature of the substance and T0 is its initial temperature.
3. The type of substance. This factor is represented mathematically through a new quantity, called specific heat capacity, c, which is the amount of heat absorbed by a 1 kg of a substance to increase its temperature by 1°C (or Kelvin).
1. Putting all the above factors together, we obtain the formula for the specific heat capacity, c:

c = Q/m × ∆T

The SI unit of specific heat capacity is [J/(kg × K)].

Rearranging the above formula, we obtain for the heat energy absorbed by an object of mass m and specific heat capacity c when its temperature increases by ΔT:

Q = m × c × ∆T

### Example 1

What is the amount of heat absorbed by 300 g water at 5°C when its temperature increases to 75°C? Take the specific heat capacity of water cw = 4186 J/kgK.

### Solution 1

We have the following clues:

m = 300 g = 0.3 kg
t1 = 5°C
t2 = 75°C
c = 4186 J/kgK
Q = ?

Applying the equation

Q = m × c × ∆T

we obtain after substitutions:

Q = m × c × (t2 - t1 )
= 0.3 kg × 4186 J/(kg × K) × (75 - 5)°C
= 0.3 kg × 4186 J/(kg × K) × 70 K
= 87 906 J

In general, dense materials such as metals, are good absorbers of heat as they contain more molecules that less dense materials and as a result, they act like reservoirs of heat energy because each molecule absorbs its own portion of energy during a heating process. Another reason why dense materials are good absorbers of heat is that their molecules can find much easier a neighbouring atom to collide with compared to less dense materials.

It is clear that vacuum is the poorest absorber of heat, as it does not contain any matter. Also, gases like air, light solids like wood, cork etc. are poor absorbers of heat.

## Joule or Calorie?

Since heat is a form of energy, its unit is normally Joule, J. However, it was not always so in the past. Initially, another unit known as calorie, cal was used as a unit of energy. It represents the amount of heat supplied to 1 g of water to increase its temperature by 1°C. Nowadays, calorie is used as a unit of energy in food industry. Since calorie is very small, a multiple of it, known as kilocalorie, kcal (or Cal with uppercase C)is used as a unit of food energy. Thus,

1 Cal = 1
kcal = 1000 cal

### Example 2

The unit calorie (cal) is based on the heat absorbed by water. Thus, 1 cal is the amount of heat energy needed to increase the temperature of 1 g water by 1 degree (Celsius or Kelvin). Based on this info, find the conversion factors between calorie and Joule given that the specific heat capacity of water is 4186 J/kgK.

### Solution 2

The value of specific heat capacity of water shows the amount of heat (in Joules) needed to increase the temperature of 1 kg water by 1 degree (Celsius or Kelvin). Therefore, given that 1 kg = 1000 g, the amount of heat energy (in Joules) needed to increase the temperature of 1 g water by 1 degree (Celsius or Kelvin) is

c(1g) = 4186 J/1000
= 4.186 J

From the info provided in the clues, this value corresponds to the heat in calories. Therefore, we can write

1 cal = 4.186 J

This number represents the conversion factor between calorie and Joule. For example, if we see on a chips package that the nutritional values per 100 g are 0.22 kCal (22°Cal), it means that if you eat 100 g of such chips, your body gets

(220 × 1000) cal × 4.186 J/cal
= 920 J

of food energy, which is partially converted into heat, in order to maintain a constant body temperature (37°C). (The rest is converted into other forms of energy needed to the body for operating regularly).

## States of Matter

Matter exists in three stable states: solid, liquid and gas. (Actually, there exist two other states of matter: plasma and Bose-Einstein particles, which are not stable, so in this tutorial we will focus only on the three abovementioned stable states.) Each of them has its own specific features which will be explained both in the visual and energetic viewpoint.

### a) Solid State

In solid state, particles are very close to each other. They are strongly bound to each other because the binding potential energy prevails over the kinetic energy of atoms. As a result, atoms can only vibrate around their equilibrium position but they cannot leave their predefined place. This means a certain atom in solid state has always the same "neighbours" around.

A cube in which some balls are placed at vertices and they are connected by each other by means of elastic springs, represents a very suitable pattern that helps us understand how solids work. Thus, if we kick one of the balls, all balls will vibrate because the energy is spread out in every ball by means of the springs. The cube can rotate as a whole, but the balls cannot move from their positions. Likewise, the particles of a solid object will vibrate if some heat energy is supplied to one part of the object. It can shift, rotate, swing etc., as a whole, but the structure of the object will remain unchanged.

The structure of a solid object looks like the one shown in the figure below. ### b) Liquid State

In liquid state, the particles slide over each other, as they are not so tightly bound. The atoms potential energy is still greater than their kinetic energy, so atoms still stay together. It is like opening a sack full of tennis-table balls and spread them throughout the room. The balls will slide over each other forming horizontal layers as shown in the figure. A liquid takes the shape of the container it is poured. Therefore, we can calculate the volume of a liquid by taking the volume of the container up to the liquid's level.

### c) Gaseous State

In gaseous state, atoms are far away from each other; they move freely in space, as the atoms kinetic energy is greater than their potential energy. A gas fills the whole space of the container it is poured, as shown in the figure below. ## Change of State

When an object changes its phase, we say there is a change of state in it. There are six possible phase changes during a change in state.

1. Melting. It occurs when a solid turns into liquid. During this process, temperature remains constant as the heat supplied to the object goes for breaking the strong molecular bonds, not for the increase in their kinetic energy.
2. Freezing. It is the reverse process of melting that occurs when a liquid turns into solid. It occurs when the object gives off heat to the surroundings. After the termination of this process, atoms are more structured than before. Again, the freeing process occurs without any change in temperature.
3. Evaporation. It occurs when a liquid turns into gas. Again, the heat supplied during evaporation does not contribute in the increase in temperature. It only allows atoms leave the liquid and move freely in space.
4. Condensation. It is the reverse process of evaporation, i.e. it occurs when a gas turns into liquid giving off heat, without any change in temperature.
5. Sublimation. Sometimes, a solid turns directly into gas due to the very large amount of heat it absorbs. During this process, the liquid phase is skipped.
6. Deposition. It is the reverse process of sublimation, i.e. it occurs when a gas turns directly into solid without passing through the liquid state. This process occurs when the object gives off large amounts of heat at a very short time.

The following scheme gives a clearer idea about the relationship between various stages in states of matter during their phase change. We use a special word to describe the process in which the evaporation takes place at the highest rate. It is known as "boiling". The boiling temperature is the highest temperature a liquid may have. For example, water boils at 100°C, which corresponds to the highest temperature of water in the liquid state.

## Latent Heat. Specific Latent Heat of Fusion and Vaporization

Since there is no change in temperature during the phase change, we cannot use anymore the formula Q = m × c × ΔT to calculate the heat supplied to or removed from a substance during such a process, as ΔT = 0 and therefore, the value of Q would result zero as well. It means we must find other ways to calculate the heat during this process.

It is clear that the substance gains or loses heat during a phase change regardless its temperature remains constant. For this reason, this kind of heat - often forgotten to be considered during calculations - is known as latent (hidden) heat.

By definition, latent heat of fusion, Qf, is the amount of heat absorbed by a substance at the melting temperature in order to melt it completely.

Similarly, latent heat of vaporization, Qv, is the amount of heat absorbed by a substance at the boiling temperature in order to evaporate it completely.

Not all substances require the same amount of heat energy to change their state. Therefore, similarly as in the case of specific heat capacity, we introduce a new quantity known as the specific latent heat, L, which represents the amount of heat per kilogram mass needed to change the phase of a material. By definition,

Specific latent heat of fusion Lf is the amount of heat supplied to 1 kg of a substance in its melting temperature in order to make it melt completely.

Similarly, specific latent heat of vaporization Lv is the amount of heat supplied to 1 kg of a substance in its boiling temperature in order to make it evaporate completely.

Both specific latent heats are measured in Joules per kilogram, [J/kg]. The equation of latent heat for both cases therefore is

Qf = m × Lf

and

Qv = m × Lv

### Example 3

What is the amount of heat absorbed by 200 g of solid lead at 600 K? Melting temperature of lead is 327°C and its specific latent heat of fusion is 22 900 J/kg.

### Solution 3

600 K correspond to 327°C because T(K) = t°C + 273°. This is equal to the melting temperature of lead. Therefore, we have to use only the equation

Qf = m × Lf

to calculate the heat absorbed by lead during this stage. Also, we have to write the mass of lead in kilograms, i.e. m = 0.3 kg. Thus, we obtain after substituting the known values

Qf = 0.3 kg × 22900 J/kg
=6370 J

for the heat absorbed by lead during the melting stage.

We can combine the formulae Q = m × c × Δ t and Q = m × L to calculate the amount of heat absorbed when the material is not at the melting or boiling temperature. Let's consider an example to clarify this point.

### Example 4

What is the amount of heat absorbed by 50 g ice at -20°C in order to melt it completely? Take the melting (freezing) temperature of ice (water) equal to 0°C. Specific heat capacity of ice is 2100 J/kgK and the latent heat of fusion for ice is 334 000 J/kg.

### Solution 4

The material passes into two stages during this event:

1. The ice is "heated" from -20°C to 0°C. During this process, we must use the formula Q = m × c × Δ T.
2. The ice is melted when it reaches the temperature 0°C. During this process, we must use the formula Q = m × Lf.

Given that m = 50 g = 0.05 kg, c = 4200 J/kgK, t1 = -20°C, t2 = 0°C and Lf = 334 000 J/kg, we write:

Qtotal = Q1 + Q2
= m × c × (t2 - t1 ) + m × Lf
= 0.05 × 2100 × [0 - (-20)] + 0.05 × 334000
= 2100 J + 16700 J
= 18800 J

The graph below represents the relationship between heat and temperature in all possible stages. In part I of the graph, the substance is in solid state. We must use the formula Q = m × cs × ΔTs (s stands for solid) to calculate the heat absorbed by it during this stage.

In part II of the graph, the substance is melting. We must use the formula Q = m × Lf to calculate the heat absorbed by it during this stage.

In part III of the graph, the substance is in liquid state. We must use the formula Q = m × cl × ΔTl (l stands for liquid) to calculate the heat absorbed by it during this stage.

In part IV of the graph, the substance is boiling (evaporating at the highest rate). We must use the formula Q = m × Lv to calculate the heat absorbed by it during this stage.

In part V of the graph, the substance is in gaseous state. We must use the formula Q = m × cg × ΔTg (g stands for gas) to calculate the heat absorbed by it during this stage.

Remark!

1. The specific heat capacities are not equal for the same material. For example, for water the value of specific heat capacity is 4186 J/kgK while for ice or water vapour, it is 2100 J/kgK. This is true for latent heats as well. For example, latent heat of fusion for ice is 334 000 J/kg and the latent heat of vaporization for water is 2 224 000 J/kg.
2. If mass is given in grams and heat in calories, we can use the conversion factor 1 cal/g × °C = 4180 J/(kg × °C) in order to deal with smaller numbers.
3. We can find the heat released by a hot object when it cools down by using the same procedure as when it is heated up. The only difference is that the heat value will result a negative number because t2 < t1 and therefore, Δt = t2 < t1 < 0.

### Example 5

A 200 ml glass full of water initially at the environment temperature (20°C), is placed inside a refrigerator whose temperature is -10°C. Given that water has a density of 1000 kg/m3 and 1 ml = 1 cm3, calculate the amount of heat (in calories) released by water during this process. Take cwater = 1 cal/gK, cice = 0.5 cal/g and Lfusion = 80 cal/g.

### Solution 5

Let's convert the clues into the required units first. We have

V = 200 ml = 200 cm3
t1 (water) = 20°C
t2 (ice) = -10°C
ρwater = 1000 kg/m3 = 1 g/cm3
cwater = 1 cal/gK
cice = 0.5 cal/g
Lfusion = 80 cal/g
Qtotal released = ?

First, we must work out the mass (in grams) of water, as it is needed during the solution. We have

m = ρ × V
= 1 g/cm 3 × 200 cm3
= 200 g

In this problem, water experiences three stages:

1) It cools down from 20°C to 0°C (the minimum temperature in which water can exist in liquid state). During this process, we must use the equation

Q1 =m × cw × ∆tw
= m × cw × (tmelting - t1(water) )
= 200 g × 1 cal/g∙°C × (0 - 20)°C
= -4000 cal

2) Then, water freezes without any change in temperature. During this process, we must use the equation

Q2 = m × Lf
= -200 g × (-80)cal/g
= -1600 cal

3) Finally, the ice cols down from 0°C to -10°C. During this process, we must use the equation

Q3 = m × cw × ∆tw
= m × cice × (t2 (ice) -tmelting )
= 200 g × 1 cal/g × °C × (-10 - 0)°C
= -2000 cal

Therefore, we obtain for the total energy

Qtotal = Q1 + Q2 + Q3
= -4000 cal - 1600 cal - 2000 cal
= -7600 cal

This means the water loses 7600 cal of heat energy during this process.

## Summary

"Heat energy always flows from the hottest object to the coldest one."

The absorption of heat energy is not equal for all materials. Some materials can absorb the heat more easily (good absorbers of heat) while some other materials show resistance to heat absorption (bad absorbers of heat).

The factors affecting the amount of heat absorbed by a substance are:

1. The amount of substance. Mathematically, this factor is represented through the mass of substance, m.
2. The increase in temperature of the substance. This factor is represented in formulae by the symbol ΔT = T - T0, where T is the final temperature of the substance and T0 is its initial temperature.
3. The type of substance. This factor is represented mathematically through a new quantity, called specific heat capacity, c, which is the amount of heat absorbed by a 1 kg of a substance to increase its temperature by 1°C (or Kelvin).

Putting all the above factors together, we obtain the formula for the specific heat capacity, c:

c = Q/m × ∆T

The SI unit of specific heat capacity is [J/(kg × K)].

Rearranging the above formula, we obtain for the heat energy absorbed by an object of mass m and specific heat capacity c when its temperature increases by ΔT:

Q = m × c × ∆T

In general, dense materials such as metals, are better absorbers of heat than less dense materials.

Initially, another unit known as calorie, cal was used as a unit of energy instead of Joule. It represents the amount of heat supplied to 1 g of water to increase its temperature by 1°C. Nowadays, calorie is still used as a unit of energy in food industry. Since calorie is very small, a multiple of it, known as kilocalorie, kcal (or Cal with uppercase C) is used as a unit of food energy. Thus,

1 Cal = 1 kcal = 1000 cal

The conversion factor between calorie and Joule is

1 cal = 4.186 J

Matter exists in three stable states: solid, liquid and gas.

1. In solid state, particles are very close to each other. They are strongly bound to each other because the binding potential energy prevails over the kinetic energy of atoms. As a result, atoms can only vibrate around their equilibrium position but they cannot leave their predefined place. This means a certain atom in solid state has always the same "neighbours" around.
2. In liquid state, the particles slide over each other, as they are not so tightly bound. The atoms potential energy is still greater than their kinetic energy, so atoms still stay together. It is like opening a sack full of tennis-table balls and spread them throughout the room. The balls will slide over each other forming horizontal layers. A liquid takes the shape of the container it is poured.
3. In gaseous state, atoms are far away from each other; they move freely in space, as the atoms kinetic energy is greater than their potential energy. A gas fills the whole space of the container it is poured

When material changes its phase, we say there is a change of state in it. There are six possible phase changes during a change in state.

1. Melting. It occurs when a solid turns into liquid. During this process, temperature remains constant as the heat supplied to the object goes for breaking the strong molecular bonds, not for the increase in their kinetic energy.
2. Freezing. It is the reverse process of melting that occurs when a liquid turns into solid. It occurs when the object gives off heat to the surroundings. After the termination of this process, atoms are more structured than before. Again, the freeing process occurs without any change in temperature.
3. Evaporation. It occurs when a liquid turns into gas. Again, the heat supplied during evaporation does not contribute in the increase in temperature. It only allows atoms leave the liquid and move freely in space.
4. Condensation. It is the reverse process of evaporation, i.e. it occurs when a gas turns into liquid giving off heat, without any change in temperature.
5. Sublimation. Sometimes, a solid turns directly into gas due to the very large amount of heat it absorbs. During this process, the liquid phase is skipped.
6. Deposition. It is the reverse process of sublimation, i.e. it occurs when a gas turns directly into solid without passing through the liquid state. This process occurs when the object gives off large amounts of heat at a very short time.

By definition, latent heat of fusion, Qf, is the amount of heat absorbed by a substance at the melting temperature in order to melt it completely.

Similarly, latent heat of vaporization, Qv, is the amount of heat absorbed by a substance at the boiling temperature in order to evaporate it completely.

Not all substances require the same amount of heat energy to change their state. Therefore, similarly as in the case of specific heat capacity, we introduce a new quantity known as the specific latent heat, L, which represents the amount of heat per kilogram mass needed to change the phase of a material. By definition,

Specific latent heat of fusion Lf is the amount of heat supplied to 1 kg of a substance in its melting temperature in order to make it melt completely.

Similarly, specific latent heat of vaporization Lv is the amount of heat supplied to 1 kg of a substance in its boiling temperature in order to make it evaporate completely.

Both specific latent heats are measured in Joules per kilogram, [J/kg]. The equation of latent heat for both cases therefore is

Qf=m × Lf

and

Qv = m × Lv

We can find the heat released by a hot object when it cools down by using the same procedure as when it is heated up. The only difference is that the heat value will result a negative number because t2 < t1 and therefore, Δt = t2 < t1 < 0.

## Physics Revision Questions for Absorption of Heat

1) How many Joules are needed to change the temperature of 200 g aluminum from 15°C to 55°C?

caluminum = 900 J/kg°C
1. 1800 J
2. 3600 J
3. 5400 J
4. 7200 J

Correct Answer: D

2) If 2000 cal is supplied to some water at 10°C, its final temperature becomes 50°C. Calculate the mass of water in grams. (cwater=1 cal/g°C)

1. 15
2. 20
3. 50
4. 85

Correct Answer: C

3) A 20 g ice cube at -15°C is converted into water at 40°C. Calculate the energy required for this process.

cice = 0.5 cal/g °C , Lice = 80 cal/g , cwater = 1 cal/g °C
1. 800 cal
2. 1600 cal
3. 2550 cal
4. 3000 cal

Correct Answer: C

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