Spread the love

Specific Heat | Specific Heat Capacity

Specific Heat | Specific Heat Capacity :- Specific heat, also known as specific heat capacity, is a thermodynamic property that quantifies the amount of heat energy required to raise the temperature of a unit mass (typically measured in grams) of a substance by one degree Celsius (or one Kelvin) under constant pressure conditions. It is denoted by the symbol “c” or “s” and is expressed in units of joules per kilogram per Kelvin, JKg^-1K^-1 or calories per gram per degree Celsius, cal. g^-1°C^-1.

The specific heat of a substance is a measure of its ability to store and release thermal energy. Different substances have different specific heat values, which depend on their molecular structure and composition. Materials with higher specific heat values require more energy to raise their temperature compared to substances with lower specific heat values.

Formula of Specific Heat

Suppose m mass of a substance is supplied ΔQ amount of heat due to which, its temperature increases by ΔT, then

Heat energy required for unit mass = ΔQ/m

⇒ Heat energy required for unit temperature change of unit mass = ΔQ/(mΔT)

Hence formula of specific heat,

$\displaystyle c=\frac{\Delta Q}{m\Delta T}$

$\displaystyle \Delta Q=mc\Delta T$

where:

$ΔQ$ is the amount of heat energy transferred (in joules)
$m$ is the mass of the substance (in grams)
$c$ is the specific heat of the substance (in JKg^-1K^-1 or cal. g^-1°C^-1)
$Δ T$ is the change in temperature (in degrees Celsius or Kelvin)

Differential form of specific heat capacity

$\displaystyle c=\frac{1}{m}\frac{dQ}{dT}$

As $\displaystyle \frac{dQ}{dT}=Heat\text{ }Capacity/Thermal\text{ }Capacity$ , so

$\displaystyle c=\frac{Heat\text{ }Capacity}{m}$

Specific heat is an essential concept in various fields, including thermodynamics, chemistry, and engineering, as it helps understand and predict how substances respond to changes in temperature and how much heat energy is needed for specific processes like heating or cooling.

The specific heat capacity of a substance depends on the following factors :

(1) Nature of the Substance – Different materials have different atomic structures, bonding, and molecular arrangements, which affect their ability to store heat. For example, water has a high specific heat capacity compared to metals.

(2) Temperature – The specific heat capacity of some substances varies with temperature. At very high or low temperatures, molecular vibrations change, affecting heat absorption.

The temperature dependence of the specific heat of water at 1 atmospheric pressure is shown in figure below. Its variation is less than 1 % over the interval from 0°C to 100°C. Such a small variation is typical for most solids and liquids, so their specific heats can generally be taken to be constant over fairly large temperature ranges.

(3) Phase of the Substance – The specific heat of a substance differs in solid, liquid, and gaseous states due to variations in molecular motion and bonding. For example, the specific heat of water in liquid form is different from that of ice or steam.

(4) Degree of Freedom of Molecules – Gases with more complex molecules (having rotational and vibrational energy states) have higher specific heat compared to monatomic gases.

(5) Pressure (for gases only) – For gases, specific heat capacity depends on whether the process is at constant volume () or constant pressure (). Due to work done by expansion, .

Example 1.

A water cooler of storage capacity $120$ liters can cool water at a constant rate of $P$ watts. In a closed circular system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly $3 kW$ of heat (thermal load). The temperature of water fed into the device cannot exceed $3 0^{\circ} C$ and the entire stored $120$ litres of water is initially cooled to $1 0^{\circ} C$ . The entire system is thermally insulated. The minimum value of $P$ (in watts) for which the device can be operated for $3$ hours is

(Specific heat of water is $4.2 k J k g^{-1} K^{-1}$ and the density of water is $1000 k g m^{-3}$ )

(A) 1600

(B) 2067

(D) 3933

[JEE (Adv.) 2016]

Solution :

Process Overview :

Device (adds heat 3 kW)
↓
Total Heat Generated in 3 h
↓
Part stored in water (raises temp 10 °C → 30°C)

↓
Remaining heat Must be removed by cooler

↓
Divide by time → Find P (cooler power)

Mass of water,

$\displaystyle m=V\rho =\left( 120\times {{10}^{-3}} \right)\times 1000=120kg$

Heat produced in 3 hr,

$\displaystyle {{Q}_{developed}}=P\times t=3000\times \left( 3\times 60\times 60 \right)=3.24\times {{10}^{7}}J$

Heat absorbed by water (from 10 °C → 30 °C),

$\displaystyle {{Q}_{water}}=mc\Delta T=120\times 4200\times (30-10)=1.008\times {{10}^{7}}J$

Remaining heat to be removed by cooler in 3 h,

$\displaystyle {{Q}_{remaining}}={{Q}_{developed}}-{{Q}_{water}}=3.24\times {{10}^{7}}J-1.008\times {{10}^{7}}J=2.232\times {{10}^{7}}J$

Minimum power of cooler,

$\displaystyle {{P}_{min}}=\frac{{{Q}_{remaining}}}{t}=\frac{2.232\times {{10}^{7}}J}{10800}\approx 2067W$

Correct option is (B).

Example 2.

A current carrying wire heats a metal rod. The wire provides a constant power (P) to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes with time (t) as :

T (t) = T₀ (1 + βt^1/4) ,

where $β$ is a constant with appropriate dimension while $T_{0}$ is a constant with dimension of temperature. The heat capacity of the metal is :

(A) $\displaystyle \frac{4P{{\left[ T(t)-{{T}_{0}} \right]}^{2}}}{{{\beta }^{4}}T_{0}^{3}}$

(B) $\displaystyle \frac{4P{{\left[ T(t)-{{T}_{0}} \right]}^{4}}}{{{\beta }^{4}}T_{0}^{5}}$

(D) $\displaystyle \frac{4P\left[ T(t)-{{T}_{0}} \right]}{{{\beta }^{4}}T_{0}^{2}}$

[JEE (Adv.) 2019]

Solution :

Heat capacity (H.C.) is give by :

$\displaystyle H.C.=\frac{dQ}{dT}$

Now

$\displaystyle \frac{dQ}{dT}=\frac{\frac{dQ}{dt}}{\frac{dT}{dt}}=\frac{P}{\frac{d}{dt}\left[ {{T}_{0}}(1+\beta {{t}^{1/4}}) \right]}$

$\displaystyle \Rightarrow \frac{dQ}{dT}=\frac{P}{\left[ {{T}_{0}}(0+\beta \times \frac{1}{4}\times {{t}^{-3/4}}) \right]}$

$\displaystyle \Rightarrow \frac{dQ}{dT}=\frac{4P}{{{T}_{0}}\beta }{{t}^{3/4}}$ …..(a)

Given that,

$\displaystyle T={{T}_{0}}(1+\beta {{t}^{1/4}})$

$\displaystyle \Rightarrow t={{\left( \frac{T-{{T}_{0}}}{\beta {{T}_{0}}} \right)}^{4}}$

Using the value of t in equation (a), we get

$\displaystyle \frac{dQ}{dT}=\frac{4P}{{{T}_{0}}\beta }{{t}^{3/4}}=\frac{4P}{{{T}_{0}}\beta }{{\left[ {{\left( \frac{T-{{T}_{0}}}{\beta {{T}_{0}}} \right)}^{4}} \right]}^{3/4}}$