# Relation Between Calorie and Joule - A Complete Guide

## What is a Calorie ?

Calorie is the unit of heat in the C.G.S. system, also called the practical unit of heat while the S.I. unit of heat being the joule. However, the Calorie unit is more widely used when heat is measured. This unit was formulated way before the time it was recognized that heat is a form of energy. In fact, the science of measuring heat transfers during a process (physical or a chemical process), i.e. how much heat has been absorbed or evolved, is called calorimetry and the apparatus which is used in measuring such heat transfers is called calorimeter. These calorimeters also help us in measuring specific heat capacities and molar heat capacities of a given material. We will discuss more about calorimeters later in the article.

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## How to Define 1 Calorie ?

We can define one calorie of heat as the amount of heat needed to raise the temperature of 1 gram of water from 14.5 ̊C to 15. 5 ̊C (i.e. increase its temperature through 1 ̊C), at a pressure of 1 atmosphere. The amount of heat needed to increase the temperature of 1g of water by 1 ̊C depends slightly on the actual temperature of water and the pressure and this is the reason why, the range 14.5 ̊C to 15. 5 ̊C and the pressure of 1 atmosphere were specified in the definition. However, we can generally define one calorie as the amount of heat needed to raise the temperature of 1 gram by 1 ̊C at any region of temperature and pressure.

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## Conversion Calorie to Joule

Now, 1 calorie is equal to how many joules? We can convert calorie to joules and vice-versa by a simple conversion relation, i.e. 1 calorie is equal to 4.186 joules.

1 Cal =4.186 joule4.2 joule

We can also use the unit kcal meaning kilocalories. Conversion of 1 kcal to cal can be done by a simple conversion relation, i.e. 1kcal=1000cal.

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## Conversion of kilograms to calories (1kg is equal to how many calories?)

We are often asked questions in our day to day lives about the amount of calories we intake from food we eat or how many calories we have burned during a workout session. This is related to body fat that is how much calories we need, to gain or lose one kg of weight. This can be found out by a simple conversion relation i.e. 1kg=7716.179176 kcal.

Example 1 - How many calories are needed to be burnt to lose 5kg weight?

Answer- 5kg= 5×7716.179176kcal= 38580.8958 kcal

Actually calories burnt in our body is the form of energy, done by working out and exercising and kg is nothing but the weight of our body which is a force. Thus kg and calorie cannot be equated. But in our daily life, we have formulated an equivalence between them based on a simple fact that the more we eat calories, the more we gain body weight and the more we workout and burn those calories, the more we lose body weight.

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## Calorimetry

Calorimetry is one of the many methods we use to determine specific heat/latent heat of a substance. Let us first define specific heat and latent heat of a substance. Specific heat or specific heat capacity can be defined as the amount of heat required to raise the temperature of the unit mass of the given substance through a unit degree.

S=ΔQ/m (ΔT), where s is specific heat measured in cal g-1℃-1 and in S.I. unit Jkg-1K-1.

Latent heat of a substance is the amount of heat needed to change the state of unit mass of a given substance from solid to liquid or from liquid to gas/vapour without any change in the temperature.

L=Q/m, where L is the latent heat measured in Jkg-1 orcal g-1.

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For example- Latent heat of fusion of ice is 80 cal g-1 and latent heat of vaporization of water is 540 cal g-1 .

Now, a calorimeter is a cylindrical vessel made up of copper provided with a stirrer and a lid. The method used for determination of specific heat/latent heat is the method of mixtures. First the calorimeter and the stirrer are cleaned, washed, dried and then weighed. Then the calorimeter is half filled with water and again its weight is recorded to measure the mass of water taken. The calorimeter is then placed in an outer jacket and we note the initial temperature of water and the calorimeter both. Now, we heat the substance whose specific heat we want to determine and put it in the calorimeter and stir the mixture. Since the substance is at higher temperature, heat is lost from it and gained by the water and the calorimeter. We again stir the mixture and note the common constant temperature of the mixture. We then weigh the mixture to find the mass of the substance added.

If w=water equivalent of calorimeter and stirrer, m1=mass of water

t1=initial temperature of water and the calorimeter, m2= mass of the given substance

s= specific heat of the substance, t2=temperature of the substance and t= common temperature of the mixture.

Now, fall in temperature of substance= (t2-t)

Rise in temperature of water and the calorimeter= (t-t1)

Heat lost by substance= s.m2.(t2-t)

Heat gained by calorimeter and water = (m1+w) (t-t1)

If there are no unwanted loss of heat, then by principle of calorimetry, heat lost= heat gained

• s.m2.(t2-t)= (m1+w) (t-t1)

• s= [(m1+w). (t-t1)]/ [m2.(t2-t)]

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