# Sensitive heat: concept, formulas and solved exercises

The **sensible heat** is the heat energy supplied to an object on its temperature subir.É the opposite of the latent heat, wherein the thermal energy does not increase the temperature, but promotes a phase change, for example, from solid to liquid.

An example clarifies the concept. Suppose we have a pot with water at room temperature of 20°C. When placed on the burner, the heat supplied increases the water temperature slowly until it reaches 100°C (water boiling temperature at sea level). The heat supplied is called sensible heat.

When the water reaches boiling temperature, the heat supplied by the burner no longer increases the water temperature, which is kept at 100 °C. In this case, the thermal energy supplied is invested in evaporating the water. The heat supplied is latent because it did not raise the temperature but caused a change from the liquid to the gaseous phase.

It is an experimental fact that the sensible heat needed to obtain a certain temperature variation is directly proportional to that variation and to the object’s mass.

__Concept and formulas__

__Concept and formulas__

It was observed that, in addition to the mass and temperature difference, the sensible heat also depends on the material. For this reason, the proportionality constant between the sensible heat and the mass product due to the temperature difference is called specific heat.

The amount of sensible heat supplied also depends on how the process is carried out. For example, it is different if the process is carried out at a constant volume than at a constant pressure.

The formula for sensible heat in an *isobaric* process , that is, constant pressure, is as follows:

*Q = cp **. **m ( **Tf **– **Ti **)*

In the equation above, *Q* is the sensible heat supplied to the object of mass *m,* which has a high initial temperature *T _{i}* towards the end of the value of

*T f*. Furthermore, it is verified in the equation

*c*

*p*is the specific heat at constant pressure of the material that the process is carried out in this way.

Also note that sensible heat is positive when absorbed by the object and causes a rise in temperature.

If heat is supplied to a gas contained in a rigid container, the process will be *isochoric,* that is, at a constant volume; and the sensible heat formula will be written like this:

*Q = **cv. **m. ( **Tf **– **Ti **)*

**The adiabatic coefficient γ**

The relationship between the specific heat at constant pressure and the specific heat at constant volume for the same material or substance is called *the adiabatic coefficient* , which is usually indicated by the Greek letter gamma γ.

The *adiabatic coefficient* is greater than unity. The heat required to raise the temperature of a body by one gram of mass by one degree is greater in an isobaric process than in an isobaric process.

This is because, in the first case, part of the heat is used to carry out mechanical work.

In addition to specific heat, a body’s heat capacity is also generally defined. This is the amount of heat needed to raise the body’s temperature by one degree Fahrenheit.

**heat capacity C**

Heating capacity is indicated with a capital *C* , specific heat with a small *c* . The relationship between the two quantities is:

*C = cm*

Where *m* is the body mass.

Specific molar heat is also used, which is defined as the amount of sensible heat needed to raise a degree of Celsius or Kelvin of temperature to one mole of substance.

**Specific heat in solids, liquids and gases**

The specific molar heat of most solids has a value close to *3* times *R* , where *R* is the universal gas constant. *R = 8.314472 J / (mol ℃)* .

For example, aluminum has specific molar heat *24.2 J / (mol ℃* ), copper *24.5 J / (mol ℃ )* , gold *25.4 J / (mol ℃ )* and sweet iron *25.1 J / (mol ) ℃)* . Note that these values are close to *3R = 24.9 J / (mol ℃)* .

On the other hand, for most gases, the specific molar heat is close to *n (R / 2)* , where *n* is an integer and *R* is the universal gas constant. The integer *n* is related to the number of degrees of freedom of the molecule that makes up the gas.

For example, in an ideal monoatomic gas, whose molecule has only three degrees of freedom of translation, the specific molar heat at constant volume is *3 (R / 2)* . But if it is an ideal diatomic gas, there are additionally two degrees of rotation, then *c **v **= 5 (R / 2)* .

In the ideal gas the following relationship between molar specific heat at constant pressure and constant volume is true: *C **P **= c **v **+ R* .

Special mention deserves water. In a liquid state at 25 ℃, water has *c **p **= 4.1813 J / (g ℃)* , water vapor at 100 degrees Celsius has *c **p **= 2,080 J / (g ℃)* and water ice a zero degree celsius has *c **p **= 2,050 J / (g ℃)* .

__difference with latent heat__

__difference with latent heat__

Matter can be found in three states: solid, liquid and gas. To change the energy state is necessary, but each substance responds to it in a different way, according to its molecular and atomic characteristics.

When a solid melts or a liquid is evaporating, the object’s temperature remains constant until all particles change state.

Therefore, it is possible that a substance is at the same time in equilibrium in two phases: solid – liquid or liquid – vapor, for example. A quantity of the substance can be passed from one state to another by adding or removing some heat, while the temperature remains fixed.

Heat supplied to a material causes its particles to vibrate faster and increase their kinetic energy. This translates to a rise in temperature.

It is possible that the energy they acquire is so great that they no longer return to their balanced position and increase the separation between them. When this happens, the temperature does not increase, but the substance changes from solid to liquid or from liquid to gas.

The heat needed for this to happen is known as *latent heat* . Therefore, latent heat is the heat by which a substance can change its phase.

Here’s the difference with sensible heat. A substance that absorbs sensible heat increases its temperature and remains in the same state.

**How to calculate latent heat?**

Latent heat is calculated by the equation:

*Q = m. L*

Where *L* can be the specific heat of vaporization or fusion. The units of *L* are energy / mass.

Scientists have given the heating various names, depending on the type of reaction in which it participates. For example, there is the heat of reaction, the heat of combustion, the heat of solidification, the heat of solution, the heat of sublimation, and many others.

The values of many of these heat types for different substances are tabulated.

**solved exercises**

**Example 1**

Suppose one that has a piece of aluminum with 3 kg of mass. Initially it is at 20°C and it is desirable to raise its temperature to 100°C. Calculate the sensible heat required.

**Solution**

First of all, we need to know the specific heat of aluminum

*c **p **= 0.897 J / (g ° C)*

So the amount of heat needed to heat the aluminum piece will be

*Q = c **p **m (Tf – Ti) = 0.897 * 3000 * (100 – 20) J*

*Q = 215280 J*

**Example 2**

Calculate the amount of heat needed to heat 1 liter of water from 25°C to 100°C at sea level. Express the result also in kilocalories.

**Solution**

The first thing to remember is that 1 liter of water weighs 1 kg, or 1000 grams.

*Q = c **p **m (Tf – Ti) = 4.1813 J / (g ℃) * 1000 g * (100 ℃ – 25 ℃) = 313597.5 J*

A calorie is a unit of energy that is defined as the sensible heat needed to raise one gram of water to one degree Celsius. Therefore, 1 calorie is equal to 4.1813 Joules.

*Q = 313597.5 J * (1 cal / 4.1813 J) = 75000 cal = 75 kcal* .

**Example 3**

A 360.16 gram piece of material is heated from 37 ℃ to 140 ℃. The thermal energy provided is 1150 calories.

Find the material’s specific heat.

**Solution**

We can write the specific heat according to the sensible heat, the mass and the temperature variation according to the formula:

*c **p **= Q / (m ΔT)*

Replacing the data, we have the following:

*c **p **= 1150 cal / (360.16 g * (140 ℃ – 37 ℃)) = 0.0310 cal / (g ℃)*

But since a calorie is equal to 4.1813 J, the result can also be expressed as

*c **p **= 0.130 J / (g ℃)*