Calorimetry: Principle, Formula, Examples, Sample Questions

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Namrata Das

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As we know that the Universe consists of matter and energy, where the motion of the molecules creates a form of energy called heat or thermal energy. Calorimetry is used to measure the amounts of heat transferred to or from a substance. In order to do so, the heat is exchanged with a calibrated object which is a calorimeter. Here, we will discuss the facts in depth. 

Key Terms: Calorimetry Principles, Law of Conservation Energy, Heat, Temperature, Change of State, Thermal Energy

Define Calorimetry?

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Just as our Universe comprises matter and energy, matter also comprises different particles like atoms and molecules, with the help of energy these particles can move invariably. So their motion is due to either colliding with each other or moving forward and backwards. The motion of these molecules and atoms makes a form of energy that is called thermal energy or heat. We all know that heat is present everywhere including the human body, volcanoes and even in the coolest places. It can be transferred from one body to another body. The flow of heat is measured by physical processes and chemical reactions. The procedure of measuring heat is defined as calorimetry. 

Calorimetry


In other words, Calorimetry can be defined as an act of qualifying change in the thermal energy of an object. The highlighting points related to calorimetry are mentioned below-

  • The temperature of a body or an object helps to identify the amount of heat present in that particular body.
  • Temperature and heat energy are directly proportionate with each other. In simple words, that means the more heat energy the more is the temperature of a body.
  • To determine the loss and gain of heat energy, the temperature of an object is measured before and after the transfer of heat takes place. Although, the difference in temperature helps to find out the heat change of a body.

Read More: Coefficient of Linear Expansion

Example of a Calorimetry

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Let’s understand this concept with an example. 

Take a cup of coffee and a bowl of chilled ice cream to keep both of them at room temperature. After an hour or 1, you will notice that coffee cools down and ice cream melts. This happens because the coffee starts releasing heat energy due to a reduction in temperature while the ice cream starts melting due to a rise in temperature as it absorbs heat from the atmosphere. 

Exothermic and Endothermic Process

It must be noted that the process of Calorimetry is possible only by using a calorimeter device. Calorimetry is the process of measuring the amount of heat transferred to or from a substance (how much amount of heat is absorbed or lost during the entire process). A calorimeter is a tool that can be used to measure the amount of heat involved in a chemical or physical process. 

Read More: Latent Heat of Water

Principle of Calorimetry

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When two bodies of different temperatures (a solid and a liquid) are kept in physical contact with each other, the process of heat transformation will start which means the body at higher temperature releases heat, while the body at a lower temperature consumes heat. This cycle goes on until it reaches thermal equilibrium. However, the principle of calorimetry determines the law of conservation energy that can be calculated by the total heat lost by the hot body is equal to the total heat gained by the cold body. 

Principle of Calorimetry

It can be written in the following manner in mathematical terms:-

Heat Lost = Heat Gained
The heat transfer in a system that can be calculated by using the below-mentioned formulae:

q=mcΔt
Whereas;

q is denoted as a measure of heat transfer;

m is denoted as the mass of the body;

c is denoted as the specific heat of the body;

Δt is denoted as a change in the temperature. 

Read More: Thermal Stress

Things to Remember

  • The principle of calorimetry: Heat lost = Heat gained.
  • Heat transfer formula: q = mcΔt where,

q is denoted as a measure of heat transfer;

m is denoted as the mass of the body;

c is denoted as the specific heat of the body;

Δt is denoted as a change in the temperature.

  • Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter).
  • Oxygen bomb calorimeters are useful in food testing laboratories to determine the amount of heat (calories) in food.
  • Change in the heat capacity of the bonds within a molecule can be measured by differential scanning calorimetry and allows researchers to detect points of enthalpy and characterize and map the temperature-specific behavior of a drug product.

Read More: Impact of temperature

Sample Questions 

Ques. If the specific heat of a substance is infinite that means 
Heat is given out
no change in temperature takes place whether heat is taken in or out of it
Heat is taken in
none of these (1 mark)

Ans. The correct answer is b 

Ques. To convert 1 gm of ice at 0 deg C into steam at 100 deg C then what amount of heat is required
100 cal
720 cal
80 cal
540 call (1 mark)

Ans. The correct answer is (b)

Explanation is mLfusion+mS(100)+mLvaporization

Ques. Calorimeter is made from which material 
brass
aluminium 
zinc 
copper (1 mark)

Ans. The Correct option is (d)

Ques. The study of Calorimetry is based on which law?
Law of Kinetic Energy 
Joule’s Law
Law of Conservation Energy 
None of these (1 mark)

Ans. The correct option is c. Law of conservation energy.

Explanation: The principle of the calorimetry states that the heat lost by the hotter body = heat gained by the colder body. Thus in calorimetry, the total heat energy of the system remains, which is the law of conservation of energy.

Ques. What is the amount of heat required to change 1g of water by 40°C. The value of C water is mentioned at 4.2 J/gm K. (2 mark)

Ans. We have available values that are as follows:- 

C= 4.2 J/gm K

m= 1g 

ΔT= 40 

We put the values according to the equation i.e., Q= mCΔT

Q= 1 X 4.2 X 40= 168 Joules

Ques. 1000 Joules of heat is applied to a mass of lead 0.5kg to change its temperature from 20°C to 40°C. Calculate its heat capacity. (2 mark)

Ans. We have available values that are as follows:- 

Q= 1000J

m= 0.5kg

ΔT= (40-20)°C= 20°C

We put the values according to the equation i.e., Q= mCΔT

1000 = 0.5 x C x.20

C = 1000/0.5 x 20

C = 100 J/kg K.

Ques. 20g at 50°C and 8 g of ice at -20°C are mixed together in calorimetry. What is the final temperature of the mixture? (2 mark)
-5°C
5°C
20°C
28°C

Ans. The correct option is d. 28°C

Let the final temperature of the mixture be t°C

Heat lost by water in calories:

H1 = 20 x 1 x (50 - t) = 1000 - 20t

Heat from ice: H2 - msΔt + mL + mswt

= 8 x 0.5 x 20 + 8 x 80 + 8 x 1 x t

H2 = 720 + 8t = 1000 - 20t

Therefore, t = 28°C

Ques. State the principle of calorimetry. (3 mark)

Ans. When two bodies of different temperatures(a solid and a liquid) are kept in physical contact with each other, the process of heat transformation will start which means the body at higher temperature releases heat, while the body at a lower temperature consumes heat. This cycle goes on until it reaches thermal equilibrium.
However, the principle of calorimetry determines the law of conservation energy that can be calculated by the total heat lost by the hot body is equal to the total heat gained by the cold body. It can be written in the following manner in mathematical terms:-

Heat Lost = Heat Gained

Ques. 1 g of ice is mixed with 100 g of water at 50°C. Find the resulting temperature of the mixture. (2 mark)

Ans. Let the temperature of water = 1000 x 1 x (50 - θ)

Heat gained by ice = 10 x 80 + 10 x 1 x θ

Now, as per the principle of calorimetry we know that, heat lost = heat gained

5000 - 1000 = 800 + 100

4200 = 100 θ

Θ = 4200/110 = 38.2°C

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CBSE CLASS XII Related Questions

  • 1.

    A circular coil of 100 turns and radius $\left(\frac{10}{\sqrt{\pi}}\right)$ cm carrying current of $5.0\,\text{A}$ is suspended vertically in a uniform horizontal magnetic field of $2.0\,\text{T}$. The field makes an angle $30^\circ$ with the normal to the coil. Calculate:

    (I) the magnetic dipole moment of the coil, and

    (II) the magnitude of the counter torque that must be applied to prevent the coil from turning.


      • 2.
        The resistance of a metal wire at \(20^\circ\text{C}\) is \(1.05\,\Omega\) and at \(100^\circ\text{C}\) is \(1.38\,\Omega\). Determine the temperature coefficient of resistivity of this metal.


          • 3.
            A current of \(4.0\,\text{A}\) flows through a wire of length \(1\,\text{m}\) and cross-sectional area \(1.0\,\text{mm}^2\), when a potential difference of \(2\,\text{V}\) is applied across its ends. Calculate the resistivity of the material of the wire.


              • 4.

                A parallel beam of monochromatic light falls normally on a single slit of width $a$ and a diffraction pattern is observed on a screen placed at distance $D$ from the slit. Explain:

                (I) the formation of maxima and minima in the diffraction pattern, and

                (II) why the maxima go on becoming weaker and weaker with its increasing number $n$.


                  • 5.

                    A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.


                      • 6.

                        A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

                          • 5 V
                          • 3.5 V
                          • 2.5 V
                          • Zero V
                        CBSE CLASS XII Previous Year Papers

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