HEAT ABSORBED: FORMULAS, HOW TO CALCULATE IT AND SOLVED EXERCISES - PHYSICAL - 2024

The absorbed heat is defined as the transfer of energy between two bodies at different temperatures. The one with a lower temperature absorbs the heat of the one with a higher temperature. When this happens, the thermal energy of the substance that absorbs heat increases, and the particles that compose it vibrate more quickly, raising their kinetic energy.

This can lead to an increase in temperature or a change in state. For example, go from solid to liquid, like ice when it melts in contact with water or soda at room temperature.

The metal spoon absorbs heat from the hot coffee. Source: Pixabay.

Thanks to heat, it is also possible for objects to change their dimensions. Thermal expansion is a good example of this phenomenon. When most substances are heated, they tend to increase in size.

An exception to this is water. The same quantity of liquid water increases its volume when it cools below 4ºC. In addition, changes in temperature can also experience changes in its density, something also very observable in the case of water.

What it consists of and formulas

In the case of energy in transit, the units of absorbed heat are 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 one dietary calorie corresponds to one kilocalorie of heat.

Calories

The calorie, abbreviated as lime, is the amount of heat necessary to raise the temperature of 1 gram of water by 1ºC.

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

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

The equivalence between units is as follows:

The problem with these older units is that the amount of heat depends on the temperature. That is, the one required to go from 70ºC to 75ºC is not the same as that required to heat the water from 9ºC to 10ºC, for example.

That is why the definition contemplates well-defined ranges: from 14.5 to 15.5 ° C and 63 to 64 ° F for the calorie and Btu respectively.

What does the amount of heat absorbed depend on?

The amount of absorbed heat that a material picks up depends on several factors:

- Mass. The larger the mass, the more heat it is able to absorb.

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

- Temperature. Adding more heat is required to obtain a higher temperature.

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

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 variation in temperature achieved by absorbing heat.

This difference has a positive sign, since when absorbing heat, it is expected that T _f > T _o. This occurs unless the substance is undergoing a phase change, such as water going from liquid to vapor. When water boils, its temperature remains constant at approximately 100ºC, no matter how fast it boils.

How to calculate it?

By putting two objects at different temperatures in contact, after a while they both reach thermal equilibrium. The temperatures then equalize and heat transfer ceases. The same happens if more than two objects come into contact. After a certain time, they will all be at the same temperature.

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

Q _absorbed = - Q _yielded

This represents an energy balance, similar to that of a person's income and expenses. For this reason, the transferred heat has a negative sign, since for the object that yields, the final temperature is lower than the initial one. Thus:

The equation Q _absorbed = - Q _yielded is used whenever two objects are in contact.

Energy balance

To carry out the energy balance it is necessary to distinguish the objects that absorb heat from those that yield, then:

Σ 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 it is necessary to know the specific heat of each participating substance. This is the amount of heat necessary to raise the temperature of 1 g of material by 1ºC. Its units in the International System are: Joule / kg. K.

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

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

250 calories are required to raise the temperature of a metal ring from 20 to 30 ºC. If the ring has a mass of 90 g. What is the specific heat of the metal in SI units?

Solution

Units are converted first:

Q = 250 calories = 1046.5 J

m = 90 g = 90 x 10 ^-3 kg

Exercise resolved

An aluminum cup 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 its final equilibrium temperature of 32 ° C. Calculate the mass of the aluminum cup, considering that there are no heat losses to the environment.

Schematic of a calorimeter. Source: Solidswiki.

Approach

As stated above, it is important to distinguish between objects that give up heat from those that absorb:

- The aluminum cup, the copper stirrer and the water absorb heat.

- The silver sample yields heat.

Data

The specific heats of each substance are supplied:

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

Solution

Silver

Q _yielded = 400 x 10 ^-3. 234 x (32 - 87) J = -5148 J

Copper stirrer

Q _absorbed = 40 x 10 ^-3. 387 x (32 - 27) J = 77.4 J

Water

Q _absorbed = 225 x 10 ^-3. 4186 x (32 - 27) J = 4709.25 J

Aluminum mug

Q _absorbed = m _aluminum. 900 x (32 - 27) J = 4500.m _aluminum

Doing use of:

Σ Q _k = 0

77.4 + 4709.25 + 4500.m _aluminum = - (-5148)

Finally the mass of the aluminum is cleared:

m _aluminum = 0.0803 kg = 80.3 g

References

1. Giancoli, D. 2006. Physics: Principles with Applications. 6 ^th. Ed. Prentice Hall. 400 - 410.
2. Kirkpatrick, L. 2007. Physics: A Look at the World. 6 ^ta Editing abbreviated. Cengage Learning. 156-164.
3. Rex, A. 2011. Fundamentals of Physics. Pearson. 309-332.
4. Sears, Zemansky. 2016. University Physics with Modern Physics. 14 ^th. Volume1. 556-553.
5. Serway, R., Vulle, C. 2011. Fundamentals of Physics. 9 ^na Cengage Learning. 362 - 374