Absorbed heat: formulas, how to calculate it and solved exercises

The absorbed heat is defined as the transfer of energy between two bodies at different temperatures. The one with the lowest temperature absorbs the heat from the one with the highest temperature. When this happens, the thermal energy of the heat-absorbing substance increases and the particles that make it up vibrate more quickly, increasing their kinetic energy.

This could result in a temperature rise or change of state. For example, from solid to liquid, such as ice when it melts in contact with water or coolant at room temperature.

Thanks to the heat, it is also possible for objects to change their dimensions. Thermal expansion is a good example of this phenomenon. When most substances heat up, they usually experience an increase in their dimensions.

An exception to this is water. The same amount of liquid water increases its volume by cooling below 4 °C. Furthermore, changes in temperature can also undergo changes in the level of its density, something also very observable in the case of water.

what is it and formulas

In the case of energy in transit, the units of heat absorbed are the Joules. However, for a long time, heat had its own units: the calorie.

Even today, this unit is used to quantify the energy content of food, although in reality, a food calorie corresponds to a kilocalorie of heat.


Calorie, abbreviated as lime , is the amount of heat needed to raise the temperature of 1 gram of water by 1°C.

In the 19th century, Sir James Prescott Joule (1818 – 1889) performed a famous experiment in which he managed to transform mechanical work into heat, obtaining the following equivalence:

1 calorie = 4,186 Joules

In British units, the heat unit is called the Btu ( British thermal unit) , which is defined as the amount of heat needed to raise the temperature of a pound of water by 1°F.

The equivalence between the units is as follows:

1 Btu = 252 calories

The problem with these older units is that the amount of heat is temperature dependent. That is, it is not the same as what is needed to go from 70 °C to 75 °C than what is needed to heat water from 9 °C to 10 °C, for example.

That’s why the definition includes well-defined ranges: 14.5 to 15.5°C and 63 to 64°F for calories and Btu, respectively.

What does the amount of heat absorbed depend on?

The amount of absorbed heat a material collects depends on several factors:

– Characteristics of the substance. There are substances that, depending on their molecular or atomic structure, are capable of absorbing more heat than others.

– Temperature More heat is needed to obtain a higher temperature.

The amount of heat, indicated as Q, is proportional to the factors described. Therefore, it can be written as:

Q = mcΔ T

Where m is the mass of the object, c is a constant called specific heat, an intrinsic property of the substance, and Δ T is the temperature change achieved by absorbing heat.

ΔT = T f – T or

This difference has a positive sign, because when absorbing heat, it is expected that f > T o. This happens unless the substance is undergoing a phase change, like water, when it changes from liquid to vapor. When water boils, its temperature remains constant at approximately 100°C, no matter how fast it boils.

How to calculate?

By putting two objects in contact at different temperatures, after some time, they both reach thermal equilibrium. Then, temperatures equalize and heat transfer ceases. The same happens if more than two objects are contacted. After a while, everyone will be at the same temperature.

Assuming that the objects in contact form a closed system, from which heat cannot escape, the principle of energy conservation applies, therefore, it can be said that:

absorbed = – Q assigned

This represents an energy balance, similar to a person’s income and expenses. That’s why the transferred heat has a negative sign, because for the object it produces, the final temperature is lower than the initial one. Therefore:

ΔT = T f – T or <0

The equation Q absorbed = – Q assigned is used whenever there are two objects in contact.

Energetic balance

To perform the energy balance, it is necessary to distinguish objects that absorb heat from those that produce:

K Q k = 0

That is, the sum of energy gains and losses in a closed system must equal 0.

the specific heat of a substance

To calculate the amount of heat absorbed, you need to know the specific heat of each participating substance. It is the amount of heat needed to raise the temperature of 1 g of material by 1 º C. Its units in the International System are: Joule / kg. K.

There are tables with the specific heat of numerous substances, usually calculated using a calorimeter or similar tools.

An example of how to calculate the specific heat of a material

It takes 250 calories to increase the temperature of a metal ring from 20 to 30 °C. If the ring has a mass of 90 g. What is the specific heat of metal in SI units?


Units are converted first:

Q = 250 calories = 1046.5 J

m = 90 g = 90 x 10 -3 kg

Exercise solved

An aluminum beaker contains 225 g of water and a 40 g copper stirrer, all at 27 °C. A 400 g sample of silver at an initial temperature of 87 °C is placed in the water.

The stirrer is used to stir the mixture until it reaches a final equilibrium temperature of 32°C. Calculate the mass of the aluminum beaker, assuming there is no loss of heat to the environment.


As stated earlier, it is important to distinguish objects that produce heat from those that absorb:

– Aluminum cup, copper stirrer and water absorb heat.

– The silver sample produces heat.


Substance-specific heatings are provided:

– Silver: c = 234 J / kg. °C

– Copper: ​​c = 387 J / kg. °C

– Aluminum c = 900 J / kg. °C

– Water c = 4186 J / kg. °C

The heat absorbed or transferred by each substance is calculated using the equation:

Q = mcλ T



yielded = 400 x 10 -3 . 234 x (32-87) J = -5148 J

copper stirrer

absorbed = 40 x 10 -3 . 387 x (32 – 27) J = 77.4 J


absorbed = 225 x 10 -3 . 4186 x (32 – 27) J = 4709.25 J

aluminum cup

absorbed = m aluminum . 900 x (32 – 27) J = 4500 .m of aluminum

Making use of:

K Q k = 0

77.4 + 4709.25 + 4500 .m aluminum = – (-5148)

Finally, the aluminum mass is eliminated:

aluminum = 0.0803 kg = 80.3 g

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