Heat Capacity and Specific Heat Capacity: Do you know that different substances respond to heat in different ways, if yes, then why? And, how can we measure the quantity of heat gained or released by an object? How can we calculate the capacity of a substance to absorb heat energy? How the amount of heat needed for a temperature change is dependent on mass and the substance involved?

Heat gain or loss result in temperature changes, state changes, or force the system to perform work. When heat is gained or lost by an object, there will be corresponding energy changes within that object. A temperature change is associated with changes in the average kinetic energy of the particles within the object. When the heat is supplied to the object, the average kinetic energy of the particles within the object increase, which cause the corresponding temperature to rise. This rise in temperature of the object depends on the amount of heat supplied and a thermal property known as heat capacity.

Heat Capacity

Heat capacity is also known as thermal capacity and it is defined as the amount of heat required to raise the temperature of a given mass of the substance by one unit without any change of phase. It tells about the capacity of a substance to absorb heat energy.

Heat capacity is an extensive property, meaning that it is dependent upon the size/mass of the sample. It means that if a sample containing twice the amount of substance as another sample then it would require twice the amount of heat energy \(\left( Q \right)\) to achieve the same change in temperature.

We can express heat capacity as:-

\(C = \frac{{\Delta Q}}{{\Delta T}}\)

Where \({\Delta Q}\) is the amount of heat supplied to the substance (of mass \(m\)) to change its temperature from \(T\) to \(T + \Delta T\).

Different substances respond to heat in different ways. The amount of rising in temperature of different substances will depend on the heat capacities of those substances. Let if a metal chair is kept in the bright sun on a hot day, it may become quite hot to the touch. An equal mass of water under the same sun exposure will not become nearly as hot. This means that water has a high heat capacity (the amount of heat required to raise the temperature of an object by \({\rm{1}}{\,^{\rm{o}}}{\rm{C}}\)). Water is very resistant to changes in temperature, while metals generally are not. So we can say that the heat-storing capacity of water is more than the same mass of iron for the same rise in temperature.

Unit of Heat Capacity

We know that the heat capacity, \(C = \frac{{\Delta Q}}{{\Delta T}}\). The \({\rm{SI}}\) unit of heat \(\left( Q \right)\) is \({\rm{J}}\) and \({\rm{SI}}\) unit of temperature is \({\rm{K}}.\) So the \({\rm{SI}}\) unit of heat capacity \(\left( C \right)\) will be \({\rm{J}}\,{{\rm{K}}^{{\rm{ – 1}}}}.\)

Specific Heat Capacity

The specific heat capacity is defined as the amount of heat absorbed or rejected by the unit mass of the substance (undergoing no physical change)  to change its temperature by one unit. Specific heat capacity is the intensive property as it is independent of the quantity or size of the matter. It depends on the nature of the substance and its temperature.

Now we can write specific heat capacity as:-

Specific Heat Capaity \( = \frac{{{\rm{ Heat}\,\rm{Capacity }}}}{{{\rm{ mass }}}}\)

Also, if \({\Delta Q}\) stands for the amount of heat absorbed or rejected by a substance of mass \(m\), when it undergoes a temperature change \({\Delta T},\) then the specific heat capacity of that substance is given by

\(S = \frac{C}{m} = \frac{1}{m}\frac{{\Delta Q}}{{\Delta T}}\)

Where,

 \(Q = \) The quantity of heat transferred to or from the object

\(m = \) The mass of the object

\(S = \) The specific heat capacity

\(C = \) Heat capacity            

\({\Delta T =}\) Change in temperature \( = \left( {{T_{{\rm{final }}}} – {T_{{\rm{initial }}}}} \right)\)

When using the above equation, the \(Q\) value can turn out to be either positive or negative. A positive \(Q\) value indicates that the object gained thermal energy from its surroundings; this would correspond to an increase in temperature and a positive \({\Delta T}\) value. A negative \(Q\) value indicates that the object released thermal energy to its surroundings; this would correspond to a decrease in temperature and a negative \({\Delta T}\) value.

\({\rm{SI}}\) unit of specific heat capacity is \({\rm{J}}\,{{\rm{kg}}^{ – 1}}\;{{\rm{K}}^{ – 1}}.\)

Difference Between Heat Capacity and Specific Heat Capacity

From the above definitions, we can see that heat capacity is an extensive property. This means that it varies on the amount of the Substance (mass of the substance). For example, \(50\) grams of iron will have a different heat capacity as \(100\) grams of the same substance. Meanwhile, specific heat capacity is an intensive property. Using the same example, \(50\) grams of iron will have the same specific heat capacity as \(100\) grams of iron.

Heat Capacity	Specific Heat Capacity
Heat capacity is the ratio of the amount of heat energy transferred to an object to the resulting increase in its temperature.	Specific heat capacity is a measure of the amount of heat necessary to raise the temperature of a unit mass of a substance by 1 degree \({\rm{K}}\).
It is an extensive property.	It is an intensive property.
Its \({\rm{SI}}\) unit is \({\rm{J}}\,{{\rm{K}}^{{\rm{ – 1}}}}\).	Its \({\rm{SI}}\) unit is \({\rm{J}}\,{\rm{k}}{{\rm{g}}^{{\rm{ – 1}}}}{{\rm{K}}^{{\rm{ – 1}}}}\).
It is a measurable physical quantity.	It is a derived quantity.
The formula for heat capacity is given by: \(C = \frac{{\Delta Q}}{{\Delta T}}\).	The formula of specific heat capacity is given by: \(S = \frac{C}{m}\).

Heat Capacity and Specific Heat Capacity – Sample Problems

Q.1 How many liters of water at \({\rm{80}}{\,^{\rm{o}}}{\rm{C}}\) should be mixed with \(40\) liters of water at \({\rm{10}}{\,^{\rm{o}}}{\rm{C}}\) to have a mixture with a final temperature of \({\rm{40}}{{\mkern 1mu} ^{\rm{o}}}{\rm{C}}\)?
Ans: From the conservation of energy, we know that

\(\sum Q = 0\)

Hence, \( \Rightarrow {Q_{{\rm{lost}}}} + {Q_{{\rm{gain}}}} = 0\)

Given:- Let \({m_1}\) liters of water at \({\rm{80}}{\,^{\rm{o}}}{\rm{C}}\) is mixed with \({m_2}\)(\(40\) liters) of water at \({\rm{10}}{\,^{\rm{o}}}{\rm{C}},\) Then by assuming no heat exchange with the environment, the heat gain by cold water will be equal to the heat lost by hot water.

From the specific heat capacity formula of that substance, we have:

\(S = \frac{1}{m}\frac{{\Delta Q}}{{\Delta T}}\)

So, the amount of heat lost by \({\rm{80}}{\,^{\rm{o}}}{\rm{C}}\) of water will be:

\({Q_{{\rm{lost }}}} = {m_1}S\left( {{T_{{\rm{final }}}} – {T_{{\rm{initial }}}}} \right)\)

\( \Rightarrow {Q_{{\rm{lost}}}} = {m_1}S(40 – 80)\)

\( \Rightarrow {Q_{{\rm{lost}}}} = \, – 40{m_1}{S}{\mkern 1mu} {\mkern 1mu} \ldots (1)\)

And, also the heat gained by \({\rm{40}}{\,^{\rm{o}}}{\rm{C}}\) of water will be :

\({Q_{{\rm{gain}}}} = {m_2}S\left( {{T_{{\rm{final}}}} – {T_{{\rm{initial}}}}} \right)\)

\( \Rightarrow {Q_{{\rm{gain}}}} = 40 \times S \times (40 – 10)\)

\( \Rightarrow {Q_{{\rm{gain}}}} = 1200S\,\,\, \ldots (2)\)

Therefore, by equating that two equations \((1)\) and \((2)\), we can find the value of \({m_1}\) (unknown quantity of water)

We have:-

\({Q_{{\rm{gain}}}} = \;- {Q_{{\rm{lost}}}}\)

\( \Rightarrow 1200S =\; – \left( { – 40{m_1}S} \right)\)

\( \Rightarrow {m_1} = 30\) litres.

Q.2. A \(12.9\,{\rm{g}}\) sample of an unknown metal at \({\rm{26}}.{\rm{5}}{\,^{\rm{o}}}{\rm{C}}\) is placed in a Styrofoam cup containing \(50\;{\rm{g}}\) water at \(88.6{\,^{\rm{o}}}{\rm{C}}.\) The water cools down and the metal warms up until thermal equilibrium is achieved at \(87.1{\,^{\rm{o}}}{\rm{C}}.\) Assuming all the heat lost by the water is gained by the metal and that the cup is perfectly insulated, determine the specific heat capacity of the unknown metal. The specific heat capacity of water is \({\rm{4}}.{\rm{18}}\frac{{\rm{J}}}{{{\rm{g}}{\,^{\rm{o}}}{\rm{C}}}}.\)
Ans: In this question, since the cup is perfectly insulated, so we can say that the quantity of the heat lost by the water (\(\left( {{Q_{{\rm{water}}}}} \right)\) equals the quantity of heat gained by the metal \(\left( {{Q_{{\rm{metal}}}}} \right)\)). 

Let first calculate the heat lost by the water:

Given:- Mass of the water, \({M_1} = 50\;{\rm{g}}\)

Specific Capacity of water. \(S = 4.18\frac{{\rm{j}}}{{{\rm{g}}{\,^{\rm{o}}}{\rm{C}}}}\)

\({T_{{\rm{initial}}}} = 88.6{\,^{\rm{o}}}{\rm{C}}\)

\({T_{{\rm{final}}}} = 87.1{\,^{\rm{o}}}{\rm{C}}\)

\(\Delta T = {T_{{\rm{final}}}} – {T_{{\rm{initial}}}} = \, – 1.5{\,^{\rm{o}}}{\rm{C}}\)

Now,

From the specific heat capacity formula, we have;

\(S = \frac{1}{m}\frac{{\Delta Q}}{{\Delta T}}\)

With the help of the above formula, let us calculate the heat lost by water \(\left( {{Q_{{\rm{water}}}}} \right)\).

We have:-

\({Q_{{\rm{water}}}} = {(mS\Delta T)_{{\rm{water}}}}\)

\(\Rightarrow {Q_{{\rm{water}}}} = \left( {50} \right)\left( {4.18} \right)\left( { – 1.5} \right)\,{\rm{J}}\)

\( \Rightarrow {Q_{{\rm{water}}}} =\; – 313.5\;{\rm{J}}\)

Here the \({\rm{ – ve}}\) sign indicates that heat is lost by the water.

Since it is given that all the heat lost by the water is gained by the metal so we can write that:

\({Q_{{\rm{matal}}}} = + 313.5\;{\rm{J}}\)

Here the \({\rm{ + ve}}\) sign indicates that heat is gained by the metal.

Now let calculate  the value of  \({S_{{\rm{metal}}}}\)

Given:- \({Q_{{\rm{metal}}}} = + 313.5\)

Mass of sample, \({M_2} = 12.9\;{\rm{g}}\)

\({T_{{\rm{initial}}}} = 26.{\rm{5}}{\,^{\rm{o}}}{\rm{C}}\)

\({T_{{\rm{final}}}} = 87.1{\,^{\rm{o}}}{\rm{C}}\)

\(\Delta T = {T_{{\rm{final}}}} – {T_{{\rm{initial}}}} = 60.{\rm{6}}{\,^{\rm{o}}}{\rm{C}}\)

Now we have:-

\({Q_{{\rm{metal}}}} = {(mS\Delta T)_{{\rm{metal}}}}\)

\(\Rightarrow 313.5 = (12.9)\left( {{S_{{\rm{metal}}}}} \right)\left( {60.6{\,^{\rm{o}}}{\rm{C}}} \right)\)

\(\Rightarrow {S_{{\rm{metal}}}} = 0.40103\frac{{\rm{j}}}{{{\rm{g}}{\,^{\rm{o}}}{\rm{C}}}}\)

\(\Rightarrow {S_{{\rm{metal}}}} = 0.40\frac{{\rm{j}}}{{{\rm{g}}{\,^{\rm{o}}}{\rm{C}}}}\).

Summary

Heat capacity is defined as the capacity of a substance to absorb heat energy or the amount of heat required to raise the temperature of a substance by one unit without any change of phase. It is an extensive property. 

We can find heat capacity as- \(C = \frac{{\Delta Q}}{{\Delta T}}\) (\({\rm{SI}}\) unit is \({\rm{J}}\,{{\rm{K}}^{{\rm{ – 1}}}}\))

Specific heat capacity is defined as the capacity of a substance to absorb heat energy or the amount of heat required to raise the temperature of one unit of substance by one unit without any change of phase. It is an intensive property.

We can calculate specific heat capacity as- \(S = \frac{C}{m} = \frac{1}{m}\frac{{\Delta Q}}{{\Delta T}}\) (\({\rm{SI}}\) unit is \({\rm{J}}\,{{\rm{kg}}^{\rm{-1}}}\,{{\rm{K}}^{{\rm{ – 1}}}}\))

Relation between heat capacity and specific heat capacity is

Specific Heat Capaity \( = \frac{{{\rm{ Heat}\,{Capacity }}}}{{{\rm{ mass }}}}\)

Water has an extremely high specific heat capacity approximately equals to \(4.2\;{\rm{J}}/{\rm{g}}{\,^{\rm{o}}}{\rm{C}}\)., which makes it good for temperature regulation.

FAQs

Q.1. What is specific heat capacity and why is it important?
Ans:  Specific heat capacity is a measure of the amount of heat energy required to change the temperature of \(1\;{\rm{kg}}\) of material by \(1\;{\rm{K}}\). Hence it is important as it will indicate how much energy will be required to heat or cool an object of a given mass by a given amount.

Q.2. What factors affect the specific heat capacity?
Ans: Specific heat capacity mainly depends on two factors
i. The change in the temperature
ii. The substance and phase of the substance

Q.3. What is the specific capacity of water?
Ans: For liquid at room temperature and pressure, the value of the specific heat capacity of water is approximately \(4.2\,{\mkern 1mu} {\rm{J}}/{\rm{g}}{\,^{\rm{o}}}{\rm{C}}.\) This implies that it takes \(4.2\) joules of energy to raise \(1\) gram of water by \(1\) degree Celsius.

Q.4. Discuss the role of the high specific heat capacity of water regarding climate in coastal areas.
Ans: The specific heat capacity of water is five times higher than the specific heat capacity of sand(or earth). Hence Sand(or earth) gets heated or cooled more rapidly as compared to water under similar conditions. Thus near the seashore there becomes a large difference in temperature between the land and sea. Due to this difference, convection air currents are set up. The cold air blows from the land towards the sea at night i.e., land breeze. Cold air blows from sea to land during the daytime, i.e., sea breeze. These breezes make the climate near the seashore moderate.

We hope you find this article on Heat Capacity and Specific Heat Capacity helpful. In case of any queries, you can reach back to us in the comments section, and we will try to solve them.