specific heat is the amount of heat necessary for it to be possible to vary the temperature of a substance or material by 1 °C. It is proportional to the amount of heat received or donated by the substance and can be calculated using a formula that involves heat, mass and temperature.
Know more: Absolute zero — the lowest theoretical temperature a body can reach
Specific heat summary
Specific heat is the amount of heat required to change the temperature of a substance or material by 1 °C.
Some factors that affect specific heat are: intermolecular forces, impurities in substances, molar mass and degrees of freedom.
Specific heat can be found through the relationship between the heat capacity and the mass of the substance.
Molar specific heat is the amount of heat per mole of substance required to change the temperature of a substance by 1°C.
Latent heat is the heat necessary to change the physical state of a substance without raising its temperature.
Sensible heat is the heat necessary to change the temperature of a substance without changing its physical state.
What is specific heat?
specific heat is the amount of Thermal energy supplied to a substance so that its temperature varies by 1 °C. All liquids, solids and gases have a specific heat for them that characterizes their behavior when they are subjected to a heat source.
this heat is proportional to that provided by the substance, so if we increase the specific heat, the amount of heat needed for the substance to change its temperature will also increase.
For example, the specific heat of aluminum is from \(0.215\ cal/g\bullet°C\), while that of water is \(1\cal/g\bullet°C\), which means that water needs to receive more heat than aluminum in order to increase its temperature. Water will also lose heat more easily than aluminum.
Specific heat table
Specific heat indicates the precise heat for a substance to vary by 1 °C, and may undergo an increase or decrease in its temperature. In the table below, we can see the specific heat values of various substances and materials.
substance or material |
specific heat (\({cal}/{g}\bullet°C\)) |
Steel |
0,1 |
fresh water |
1 |
Salty water |
0,93 |
Ethyl alcohol |
0,58 |
Aluminum |
0,215 |
Air |
0,24 |
Sand |
0,2 |
Carbon |
0,12 |
Lead |
0,0305 |
Copper |
0,0923 |
Ethanol |
0,58 |
Iron |
0,11 |
Ice (-10°C) |
0,53 |
Granite |
0,19 |
Hydrogen |
3,4 |
Brass |
0,092 |
Wood |
0,42 |
Mercury |
0,033 |
Nitrogen |
0,25 |
Gold |
0,03 |
Oxygen |
0,22 |
Silver |
0,0564 |
Tungsten |
0,0321 |
Glass |
0,2 |
Specific heat formula
We can calculate the specific heat using the formula for the amount of heat, represented below:
\(c=\frac{Q}{m∙∆T}\)
ç → specific heat, measured in \([J/(kg\bullet K)]\) or \([cal/g\bullet°C]\).
Q → amount of heat, measured in Joule [J] or calories [cal].
m → mass, measured in kilogram [kg] or gram [g].
\(∆T \) → temperature variation, measured in Kelvin [K] or Celsius [°C].
THE temperature variation can be calculated using the following formula:
\(∆T=T_f-T_i\)
\(∆T\) → temperature variation, measured in Kelvin [K] or Celsius [°C].
\(T_f \) → final temperature, measured in Kelvin [K] or Celsius [°C].
\(You\) → initial temperature, measured in Kelvin [K] or Celsius [°C].
Important: Although the above quantities are measured in Joule, kilogram and Kelvin in the International System of Units (YES), it is more common to use calorie, gram and Celsius. It is possible to convert calorie to Joule, considering that 1 cal is equivalent to 4.186 J.
To convert grams into kilograms, just remember that 1 g is equal to 0.001 kg. In addition, to transform Celsius into Kelvin, just add to the temperature of Celsius the value of 273.15, that is, 100 °C = 373.15 K.
How to calculate specific heat?
Specific heat can be calculated using the formula that relates it to the amount of heat, mass and temperature of the substance or material.
Example:
What is the specific heat of an object with a mass of 100 g that received 1000 cal and had its temperature varied by 15 °C until reaching 35 °C?
Resolution:
like all measurement units are in their most common form, there is no need for conversion. We will use the formula for specific heat, which involves heat, mass and temperature:
\(c=\frac{Q}{m∙∆T}\)
\(c=\frac{Q}{m\bullet (T_f-T_i)}\)
Substituting the values given in the statement, we have:
\(c=\frac{1000}{100\bullet (35-15)}\)
\(c=\frac{1000}{100\bullet (20)}\)
\(c=\frac{1000}{2000}\)
\(c=0.5\)
Therefore, the specific heat of the object is\(0.5\cal/g\bullet°C\).
Factors that affect specific heat
There are a few factors that can affect specific heat variations. See below.
intermolecular forces: The specific heat varies in proportion to the intermolecular strength of the molecule, and the greater the bond, the greater the energy required to break it. Typically, molecules containing hydrogen bonds are those that contain high values of specific heat.
Impurities: The specific heat may vary with the amount of impurities in the material, even though these impurities are necessary for the formation of the material.
Molar mass: The specific heat can also vary according to the molar mass of the substance.
Degrees of freedom: The molar specific heat, as we studied in Thermodynamics, varies according to the degrees of freedom of a molecule. Briefly, it is about the freedom of movement of a molecule, and it can have translation, rotation and oscillation movements.
Specific heat and heat capacity
Also called heat capacity, heat capacity is a proportionality constant that relates the heat received or lost by a body to its temperature variation. It is possible to calculate the specific heat through the heat capacity and the mass of the substance or material with the formula:
\(c=\frac{C}{m}\)
ç → specific heat, measured in \([J/kg\bullet K]\) or \([cal/g\bullet°C]\).
Ç → heat capacity, measured in \([J/K]\) or \([cal/°C]\).
m → mass, measured in kilogram [kg] or gram [g].
Also know: Thermal expansion of solids — the phenomenon resulting from the increase in temperature of a body
molar specific heat
The molar specific heat expresses the amount of specific heat of a substance in mole, unlike specific heat, where the amount of substance is expressed in kilograms. Since we work with molecules, the size of which is tiny, it is more advantageous to express their quantity in moles than in kilograms or other units.
\(1\ mol=6.02\times{10}^{23}\ units\ elementary\ of\ any\ substance\)
For example, 1 mole of aluminum is equivalent to \(6.02\times{10}^{23}\) aluminum atoms.
The formula for calculating the molar specific heat is the same as that used for calculating the specific heat, but they differ in the unit of measurement — for the molar specific heat, use \(cal/mol\bullet°C\).
Latent heat and sensible heat
Heat can be classified as latent or sensitive. See below.
→ latent heat
O latent heat is that necessary to change the physical state of a substance without raising its temperature. It can be calculated by the formula:
\(Q=m\bullet L\)
Q → amount of heat, measured in Joule [J] or calories [cal] .
m → mass, measured in kilogram [kg] or gram [g] .
L → latent heat, measured in \([J/kg]\) or \([cal/g]\).
→ sensible heat
Sensible heat is the heat required to change the temperature of a substance without changing its physical state. It can be calculated by the formula:
\(Q=m\bullet c\bullet∆T\)
Q → amount of heat, measured in Joule [J] or calories [cal] .
m → mass, measured in kilogram [kg] or gram [g].
ç → specific heat, measured in \([J/(kg\bullet K)]\) or \([cal/g\bullet°C]\).
∆T → temperature variation, measured in Kelvin [K] or Celsius [°C].
Solved exercises on specific heat
question 1
(UFPR) To heat 500 g of a certain substance from 20 °C to 70 °C, 4000 calories were needed. The heat capacity and specific heat are, respectively:
A) 8 cal/°C and 0.08 \(\frac{cal}{g\ °C}\)
B) 80 cal/°C and 0.16 \(\frac{cal}{g\ °C}\)
C) 90 cal/°C and 0.09 \(\frac{cal}{g\ °C}\)
D) 95 cal/°C and 0.15 \(\frac{cal}{g\ °C}\)
E) 120 cal/°C and 0.12 \(\frac{cal}{g\ °C}\)
Resolution:
Alternative B
We will find the value of the heat capacity using the formula:
\(C=\frac{Q}{∆T}\)
\(C=\frac{4000\ }{70-20}\)
\(C=\frac{4000\cal}{50}\)
\(C=80\cal/°C\)
Finally, we will calculate the value of the specific heat:
\(4000=500\bullet c\bullet50\)
\(4000=25000\bullet c\)
\(\frac{4000}{25000}=c\)
\(0.16\frac{cal}{g °C}=c\)
question 2
(PUC-RS) A homogeneous body A, mass 200 g, changes its temperature from 20 °C to 50 °C when receiving 1200 calories from a thermal source. During the entire warm-up, body A remains in the solid phase. Another homogeneous body B, consisting of the same substance as body A, has twice its mass. What, in cal/g°C, is the specific heat of the substance of B?
A) 0.1
B) 0.2
C) 0.6
D) 0.8
E) 1.6
Resolution:
Alternative B
We will calculate the specific heat of material A using the sensible heat formula:
\(Q=m\bullet c\bullet\mathrm{\Delta T}\)
\(1200=200\bullet c\bullet (50-20)\)
\(1200=200\bullet c\bullet30\)
\(1200=6000\bullet c\)
\(c=\frac{1200}{6000}\)
\(c=0.2\ cal/g°C\)
The specific heat of body B will have the same value as the specific heat of body A, since they are made up of the same substance.
By Pâmella Raphaella Melo
Physics teacher
Source: Brazil School - https://brasilescola.uol.com.br/fisica/calor-especifico.htm