Ratio of Specific Heats: Specific heat capacity is the amount of heat required to raise the temperature of the unit mass of a substance through one degree. 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. The ratio of specific heat capacity to molar heat capacity of a body depends on the molecular weight of the body.

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. In this article, let us discuss the ratio of specific heat in different substances, and a few problems to solve to prepare for your examination.

What is 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 \(T + \Delta T\).

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 \({\rm{J}}\,{{\rm{kg}}^{ – 1}}\;{{\rm{K}}^{ – 1}}.\)

Specific Heat of Water

The specific heat capacity of water at normal pressure and temperature is approximately 4.2 J ⁄ g °C or 1 Cal ⁄ g °C. This means that 1 gm of water requires 4.2 joules of energy to raise 1 degree Celsius. This number is actually pretty high. Even water vapour has a higher specific heat capacity than many other materials at normal temperature. Water vapour’s specific heat capacity at normal pressure and temperature is approximately 1.9 J ⁄ g °C.

Water’s temperature falls as it releases heat and rises as it absorbs heat, as it is with other liquids. But the liquid water temperature falls or rises slower than that of many other liquids. We may conclude that water absorbs heat without causing an instant temperature increase. It also keeps its temperature for a considerably longer period of time than many other substances.

We employ this property of water in the human body to keep it at a stable temperature. There would be a lot more instances of underheating and overheating if water had a lower specific heat value.

Specific Heat at Constant Pressure or Volume

When a solid is heated over a limited temperature range, its pressure remains constant. At constant pressure, this is referred to as specific heat, abbreviated as C_P.

When a solid is heated over a short temperature range, its volume remains constant. At constant volume, this is referred to as specific heat, abbreviated as C_V.

The way gas is heated affects the behaviour of the gas, the volume and pressure change in temperature, and the amount of heat necessary to increase the temperature of 1gm of gas by 1° C. We can heat the gas with a variety of P and V values.

As a result, the specific heat value is limitless. If we don’t deliver a steady quantity of heat, the gas’s specific heat will change. As a result, we will need a constant pressure or volume of specific heat.

For an ideal gas,

C_P – C_V = n R

where, C_V is heat capacity at constant volume, C_P is heat capacity at constant pressure, R is the molar gas constant, and and n is amount of substance.

The value of gas constant, R = 8.3145 J mol^-1 K^-1

C_P ⁄ C_V Ratio (Heat Capacity Ratio)

The adiabatic index is also known as heat capacity ratio or ratio of specific heat capacities (C_P : C_V) in thermodynamics. The ratio of heat capacity at constant pressure (C_P) to heat capacity at constant volume (C_V) is defined as heat capacity ratio.

The isentropic expansion factor, commonly known as heat capacity ratio, is indicated by γ for an ideal gas (gamma). As a result, specific heat ratio, γ is equal to ratio of C_P to C_V, i.e. γ = C_P ⁄ C_V.

Why is C_P Greater than C_V?

The specific heats of an ideal gas are represented by C_P and C_V. This is the amount of heat required to raise the temperature of unit mass by 1° C. By the first law of thermodynamics,

ΔQ = ΔU + ΔW

where, ΔQ is the amount of heat that is given to the system, ΔU is the change in internal energy, and ΔW is the work done.

At constant pressure, heat is absorbed to raise internal energy and do any work on the system. On the other side, heat is absorbed just to raise internal energy at constant volume, not to do any work on the system. As a result, the specific heat under constant pressure is greater than that at constant volume, i.e. Cp > Cv.

Ratio of Specific Heats – 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:-

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

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

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

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

We hope this article on Ratio of Specific Heats has provided you with all the information and problems related to the chapter. Stay tuned to embibe.com for more of such informative articles.