The Universe is made up of two elements: **Matter **and **Energy**. The matter is made up of atoms and molecules, and energy causes these atoms and molecules to move constantly â€“ either by vibrating back and forth or colliding with one another. This movement of molecules and atoms generates a type of energy known as **Thermal Energy** or **Heat**. Heat exists in all materials, even the coldest gaps of space. This article discusses **Calorimetry**, which is a way of measuring heat transfer that occurs within a chemical reaction or other physical processes. Before jumping to the main topic firstly let’s understand what is Heat and Temperature.

**Heat: **When two bodies having different temperatures are kept in contact, the body having a higher temperature cools down and the body having a lower temperature warms up. Here the energy is transferred when the two bodies having temperature differences come in contact with each other. Hence, the energy which is transferred between two bodies as a result of temperature difference is known as **Heat**. The SI unit of heat is Joule (J).

**Temperature: **The physical quantity of a material which expresses the hotness or coldness of a material, is known as **Temperature**. The temperature is measured using a thermometer or calorimeter. Temperature is measured by three scales Celsius, Fahrenheit, and Kelvin. The SI unit of temperature is Kelvin (K).

### Thermal Properties of Matter

The physical properties of matter, which are displayed by a material when the heat is passed through it, are known as thermal properties of matter. These are the properties that decide the nature of the matter in the presence of heat, i.e. how the material will behave when it is subjected to either excessive heat or low heat. The thermal properties of matter are related to its conductivity of heat. There are four kinds of thermal properties of the matter discussed below:

**1. Heat Capacity: **The amount of heat that is required to raise the temperature of the material by 1 Â°C without any change of phase, is known as **Heat Capacity.** This amount of heat is generally expressed in Joules or Calories and the temperature fluctuations or simply temperature is expressed in Celsius or Fahrenheit. The SI unit of heat capacity is Joule per Kelvin (J/K).

- The
**Specific Heat Capacity**of a material is the amount of heat that is required to raise the temperature of 1 gram of a material by 1 Â°C. - The
**Molar Heat Capacity**of a material is the amount of heat that is required to raise the temperature of 1 mole of a material by 1 Â°C.

**2. Thermal Expansion: **When heat is passed through a material, its shape changes from its original. In other words, an object expands on heating. So there is an increase in the size of the body due to an increase in temperature. This property of the material is termed **Thermal Expansion**. Thermal expansion is measurably high in the case of gases, while in the case of liquids and solids it is relatively small.

**3. Thermal Conductivity: **Thermal conductivity is the property of a material to transfer/conduct heat. Thermal conductivity occurs when two objects are placed in contact. These materials are called **conductors**. Among the conductors, heat moves from the materials having high thermal energy to the materials having low thermal energy. This transfer of heat will continue until the thermal equilibrium is maintained between the objects kept in contact with each other. Below is the image showing thermal conductivity.

The materials which do not conduct heat at all, are known as insulators. These objects show insulating properties. For example, normal window glass will conduct less heat than an iron rod.

The SI unit of thermal conductivity is W/m.K (Watts per meter-Kelvin). Thermal Conductivity can be calculated using the formula given below:

**k = Q Ã— L/A (T _{2} – T_{1})**

where k is the thermal conductivity, Q is the heat flow, L is the thickness or length of the object, A is the surface area of the object and (T_{2} – T_{1}) is the temperature gradient.

**4. Thermal Stress: **Due to thermal expansion or contraction the body experiences some kind of stress. This is called thermal stress. The disadvantage of thermal stress is that it has the potential to destroy the object as it can make the material explode, so it may turn out to be destructive in nature. Apart from this disadvantage, thermal stress can also have advantages, for example, while joining two parts by heating the one part in manufacturing and then sliding it over the other which allows this combination to cool.

The formula to calculate thermal stress is as below:

**Î´ _{t} = Î± Ã— L Ã— (T – T_{0})**

where Î´ is the deformation due to temperature change, Î± is the temperature coefficient of expansion, L is the original length, T is the final temperature and T_{0} is the initial temperature.

For example, cracks can be seen on the tires of big trucks, driving at high speeds on the road causes friction between the road’s surface and the tires that produce heat resulting in thermal expansion. As a result of this thermal expansion, the tires experience stress. So the crack is a result of thermal stress. So, let us now study Calorimetry which is explained below,

### Calorimetry

Calorimetryis the branch of physics that monitors changes in a body’s heat energy. We are all aware that heat is a type of energy. The temperature of a body indicates the amount of heat it contains. As a result, the greater the temperature, the larger the heat energy of a body.

As a result, to determine whether a body has received or lost heat energy, we measure the temperature of the body before and after the transfer. The change in heat energy of the body is determined by the variation in temperature.

Calorimetry is the act or science of observing the change in a body’s state variables in order to measure the heat transfer associated with changes in its states such as physical changes or phase transitions under certain conditions. A **calorimeter **is used to perform calorimetry. Keep in mind that this topic mainly covers the transmission and conversion of â€˜Heat’ energy into other kinds of energy such as work and vice versa.

### Calorimeter

A device that is used for the measurement of heat (thermal changes of a body) necessary for calorimetry, is known as

Calorimeter.

A simple calorimeter is a vessel that is generally made up of copper with a stirrer of the same material. The vessel is kept in a wooden box to separate it thermally from the surrounding. A thermometer is used for the measurement of the temperature of the contents of the calorimeter. There is one opening through which a thermometer can be inserted to measure the change in thermal properties inside.

Objects which are at different temperatures are made to come in touch with each other in the calorimeter. In a calorimeter a fixed amount of fuel is burned, leading to the heating of water. Heat which is lost by the fuel is equal to the heat which is gained by the water. As a result, heat is exchanged between the objects and the calorimeter. Neglecting any heat exchange with the surrounding. This is why it is important that the calorimeter should be insulated from the environment; to improve the accuracy of the experiment. This exchange of heat is measured through the thermometer.

### Calorimeter Principle

When two bodies of different temperatures (ideally a solid and a liquid) come into physical contact with one other, heat is transferred from the body with higher temperature to the body with lower temperature until thermal equilibrium is reached between them. The body at higher temperatures emits heat, whereas the body at lower temperatures absorbs heat.

The principle of calorimetry explains the law of conservation of energy, which states that the total heat lost by the hot body equals the total heat gained by the cool body, that is:

Heat Lost = Heat GainedThe heat is calculated using the Calorimetry formula,

Q = mCâˆ†Twhere Q is the heat capacity, m is the mass, C is the specific heat capacity and Î”T is the temperature change.

**Latent Heat**

The quantity of heat required to change the unit mass of substance completely from one state into another state at a constant temperature is known as **latent heat**. However, the amount of heat required to change the unit mass of solid into its liquid state at its melting point at a constant temperature, is known as **latent heat of fusion**. For example; the Latent heat of fusion of ice is 80 kcal kg^{-1}.

The amount of heat that is needed to melt a solid of mass m may be written as:

**Q = mL**

where L is a constant for the given material (and surrounding conditions) or known as the specific latent heat of fusion/vapourisation.

This equation is also valid when a liquid changes its phase to vapor. Here, the constant L is called the specific latent heat of fusion, or Latent heat of fusion is also used to mean the same thing.

When a solid melts, then molecules of this melted solid move apart against the strong molecular attraction and this needs some energy which is supplied from outside. Thus, the internal energy of a given material is larger in the liquid phase than that in the solid phase. Similarly, in vapor phase, the internal energy of a given material is larger than that in the liquid phase.

The amount of heat required to change the unit mass of liquid to its vapor state at its boiling point at a constant temperature is known as **latent heat of vaporization**. For example; the latent heat of vaporization of H_{2}O is 540 kcal kg^{-1}.

### Sample Problems

**Problem 1: A person’s skin is more severely burnt when put in contact with 1 g of steam at 100 ^{o}C than when put in contact with 1 g of water at 100 ^{o}C. Explain.**

**Solution:**

Due to latent heat of vaporization, steam has greater energy than that boiling water. The internal energy of boiling water is at 100

^{o}C is less than the internal energy of vapor at the same temperature. Therefore, a person’s skin is more severely burnt when put with 1 g of steam at 100^{o}C.

**Problem 2: A calorimeter is kept in a wooden box to insulate it thermally from the surroundings. Why is it necessary? **

**Solution:**

If a calorimeter is not kept in an insulated wooden box then the heat of the calorimeter must be exchanged with the surrounding and this violates the principle of calorimeter because to determine the exact specific heat capacity, the total transferred heat must be known. Therefore, a calorimeter is kept in a wooden box to insulate it thermally from the surroundings.

**Problem 3: When a solid melt or a liquid boils, the temperature does not increase even when heat is supplied. Where does the energy go?**

**Solution :**

When a solid melt or a liquid boils, the temperature does not increase even when heat is supplied because the heat which is supplied is actually used to break the bond of forces between the molecules of the material and bring the molecules apart until the body completely changes its state. Hence, this supplied heat is transferred to the molecules as kinetic energy and the temperature of the material remains constant during the process.

**Problem 4: An aluminium vessel having a mass of 0.5 kg contains 0.2 kg water at 20 ^{o}C. A block of iron of mass 0.2 kg at 100 ^{o}C is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron, and water are 910 J kg^{-1} K^{-1}, 470 J kg^{-1} K^{-1}, and 4200 J kg^{-1} K^{-1} respectively.**

**Solution:**

Given,

Mass of aluminium vessel = 0.5 kg

Mass of water = 0.2 kg

Mass of block of iron = 0.2 kg

Temperature of water and aluminium = 20

^{o}C = 20 + 273 = 293KTemperature of iron = 100

^{o}C = 100 + 273 = 373KSpecific heat of aluminium = 910 J kg

^{-1}K^{-1}Specific heat of water = 4200 J kg

^{-1}K^{-1 }Specific heat of iron = 470 J kg

^{-1}K^{-1}Let T be equilibrium temperature of mixture.

Heat gained by water = 0.2 Ã— 4200 Ã— (T-293)

Heat gained by iron = 0.5 Ã— 910 Ã— (T-293)

Therefore,

Total heat gained, H

_{1}= 0.2 Ã— 4200 Ã— (T-293) + 0.5 Ã— 910 Ã— (T-293)= (T-293)[0.2 Ã— 4200 + 0.5 Ã— 910]

= (T-293)[840 + 455]

= (T-293)1295

Heat lost by iron, H

_{2}= 0.2 Ã— 470 Ã— (373-T)= 94 (373-T)

Now, we know that,

Heat gained = Heat lost

(T-293) 1295 = 94 (373-T)

(T-293)1295 / 94 = (373-T)

(T-293)14 = (373-T)

14T – 4102 = 373 – T

15T = 4475

T = 4475/15 = 298.33K â‰ˆ 298K

Therefore, T = (298 – 273) = 25

^{o}CHence, the equilibrium temperature of the mixture is

25^{o}C.

**Problem 5: A cube of iron (Density = 8000 kg m ^{-3}, specific heat capacity = 470 J kg^{-1}K^{-1}) is heated to a high temperature and is placed on a large block of ice at 0^{o}C. The cube melts the ice below it, displaces the water**,

**and just goes inside the ice. Calculate the initial temperature of the cube. Neglect any loss of heat outside the ice and the cube. The density of ice = 900 kg m**

^{-3}and the latent heat of fusion of ice = 3.36 Ã— 10^{5}J kg^{-1}.**Solution:**

Given:

Density of iron cube = 8000 kg m

^{âˆ’3}Specific heat capacity, s = 470 J kg

^{âˆ’1}K^{âˆ’1}Density of the ice = 900 kg m

^{âˆ’3}Latent heat of fusion of ice, l = 3.36 Ã— 10

^{5}J kg^{âˆ’1}Let, v be the volume of cube.

Volume of water displaced = v

Mass of cube, m = 8000 v kg

Mass of the ice melted, M = 900 v

Let, t K be the initial temperature of the iron.

Now, We know

Heat gained = heat lost

Therefore,

Heat gained by the ice = Heat lost by the iron cube

m Ã— s Ã— (t âˆ’ 273) = M Ã— l

â‡’ 8000 V Ã— 470 Ã— (T âˆ’ 273) = 900 VÃ—( 3.36 Ã— 10

^{5})â‡’ 376 Ã— 10

^{4}Ã— (T âˆ’ 273) = 3024 Ã— 10^{5}â‡’ 376 Ã— (T âˆ’ 273) = 30240

â‡’ 376T – 102648 = (30240 + 102648)/376

â‡’ T = 132888/376

â‡’ T = 353.43 K â‰ˆ 353 K

â‡’ T = 353 – 273 = 80

^{o}CHence, the initial temperature of the cube is 80

^{o}C.

**Problem 6: Calculate the time required to heat 20 kg of water from 10 ^{o}C to 35^{o}C using an immersion heater rated 1000 W. Assume that 80% of the power input is used to heat the water. Specific heat capacity of water = 4200 J kg^{-1} K^{-1}.**

**Solution:**

Given that,

Mass of water, m = 20 kg

Change in temperature, Î”t = 25 Â°C

Power rating of immersion rod, p = 1000 W

Specific heat capacity of water, s = 4200 J kg

^{âˆ’1}K^{âˆ’1}Total amount of heat required to raise the temperature from 10Â°C to 35Â°C,

Q = m Ã— s Ã— Î”t

Q = 20 Ã— 4200 Ã— 25

Q = 20 Ã— 4200 Ã— 25 = 21 Ã— 10

^{5}JLet, t be the time taken to heat 20 kg of water from 10Â°C to 35Â°C.

Only 80% of the power input is used to heat the water. Therefore,

Energy of the immersion rod used for heating water = t Ã— 0.80 Ã— 1000 J

Now,

t Ã— 0.80 Ã— 1000 = 21 Ã— 10

^{5}t Ã— 800 = 21Ã—10

^{5}=2625 sâ‡’t=262560=43.75 min â‰ˆ44 mint = (21 Ã— 10

^{5})/800= 21000/8 = 2625 s = 43.75 min

= 43.75 min â‰ˆ 44 min

Hence, the required time is

44 min.

**Problem 7: A brick weighing 4 kg is dropped into a 1 m deep river from a height of 2 m. Assuming that 80% of the gravitational potential energy is finally converted into thermal energy, find this thermal energy in calories. **

**Solution:**

Given:

Mass of brick, m = 4 kg

Total vertical distance covered by brick, h = 3 m

Gravitational potential energy converted into thermal energy = 80%

Therefore,

Change in potential energy of brick = mgh = 4 Ã— 10 Ã— 3 = 120 J

Now,

Thermal Energy = 120 Ã— (80/100) = 96 J

Thermal energy in calories, T = 96/4.2

= 22.86 cal â‰ˆ 23 Cal

Hence, the required thermal energy in calorie is

23 Cal.