# specific heat capacity formula class 11 si unit of water derivation in physics with time

find specific heat capacity formula class 11 si unit of water derivation in physics with time ?

If the temperature of a body is increased, its length, surface area and volume all increase. (i) Coefficient of Linear Expansion If the body is in the form of a rod, the increase △L in its length when the temperature is increased by △T is proportional to (a) the original length L and (b) increase in temperature △T, i.e.

where α is the coefficient of linear expansion of the material of the body. Thus

(ii) Coefficient of Area Expansion

(v) Variation of Density with Temperature

Thus density of a substance decreases with increase in temperature.

(vi) Thermal Stress If a rod is held between two rigid supports and its temperature is increased or decreased, the rigid supports prevent the rod from expanding or contracting. As a result, a stress (called thermal stress) is developed in the rod. The change in length of the rod is

**Measurement of Heat**

Heat is a form of energy. It is, therefore, measured in energy units. The SI unit of heat is joule (J). Another unit commonly used is the calorie. A calorie is the amount of heat required to raise the temperature of 1 g of water through 1°C. Experiments have shown that 4.18 J of mechanical work produce one calorie of heat.Thus

1 calorie = 4.18 joules

or 1 cal = 4.18 J

If two substances having different temperatures are brought into thermal contact, heat energy will flow from the hotter to the colder substance. The flow of heat energy will continue till the temperatures are equalized. Wesay that the thermal equilibrium has been attained. The common temperature at thermal equilibrium is called the equilibrium temperature. If no heat energy is allowed to escape to the surroundings, the amount of heat energy gained by the initially colder body is equal to the amount of heat energy lost by the initially hotter body, i.e. Heat gained by one body = heat lost by the other body. Thus is the basic principle of calorimetry. The heat energy Q needed to raise the temperature through △T of a mass m of a substance of specific heat capacity s is given by

Q = ms △T

This is the basic heat formula.

**Specific heat capacity**

If m = 1 unit and △T = 1 unit, then s = Q. Hence the specific heat capacity of a substance is the amount of heat required to raise the temperature of a unit mass of the substance through a unit degree. A commonly used unit of s is cal g^{–1} °C^{–1}. In the SI system s is expressed J kg^{–1} K^{–1}. The two units are related as

Since the size of a degree on the celsius scale is equal to that on the kelvin scale, a temperature difference of, say, 1 °C is equal to a temperature difference of 1 K. Thus

1 cal g^{–1} °C^{–1} = 4180 J kg^{–1} K^{–1}

The specific heat capacity of water = 1 cal g^{–1} °C^{–1} or 4180 J kg^{–1} K^{–1}.

**Molar specific Heat**

The molar specific heat C of a substance is the amount of heat energy required to raise the temperature of 1 mole of the substance through 1 K. It is expressed in J mol^{–1} K^{–1}. Specific heat (s) and molar specific heat (C) are related as

s = c/m

where m is the number of kilograms per mole in the substance.

**Two Specific Heats of a Gas**

Two specific heats of a gas are of special significance, namely, the specific heat at constant value (Cv) and that at constant pressure (Cp). They are defined as: The molar specific heat of a gas at constant volume (Cv) is the amount of heat energy required to raise the temperature of 1 mole of the gas through 1 K when its volume is kept constant. The molar specific heat of a gas at constant pressure (Cp) is the amount of heat energy required to raise the temperature of 1 mole of the gas through 1 K when its pressure is kept constant.

The two specific heats of an ideal gas are related as

Cp – Cv = R

when R is the universal gas constant and its value is R = 8.315 J mol^{–1} K^{–1}

**Latent Heat**

The heat energy supplied to a substance to change from solid to liquid state or from liquid to gaseous state is not registered by a thermometer as the heat energy is used up in bringing about a change of state. Hence it is called latent (or hidden) heat. A substance has two latent heats. 1. The latent heat of fusion (or melting) of a substance is the heat energy required to convert a unit mass of a substance from the solid to the liquid state, without change of temperature. The latent heat of fusion of ice is 3.36 x 105 J kg^{–1} or 80 cal g^{–1}. 2. The latent heat of vaporisation (or boiling) of a substance is the heat energy required to convert a unit mass of the substance from the liquid to the gaseous state, without change of temperature. The latent heat of vaporisation of steam is 2.26 x 106 J kg^{–1} or 540 cal g^{–1}. The amount of heat energy required to change the state of a mass m of a substance without change in temperature is given by

Q = mL

where L is the latent heat

If a steady temperature difference (T_{1} – T_{2}) is to be maintained between the ends of a rod, heat energy must be supplied at a steady rate at one end and the same must be taken out at the other end. Suppose heat energy Q flows through a rod in time t, then the rate of flow of heat Q/t through the rod in the steady state is

(i) proportional to cross-sectional area A of the rod,

(ii) proportional to temperature difference (T_{1} – T_{2}) to be maintained between the ends of the rod and

(iii) inversely proportional to length L of the rod.

where k is a constant called thermal conductivity of the material of the rod. It is a measure of how quickly heat energy is conducted (or transferred) through the substance. The SI unit of k is J s^{–1} m^{–1} K^{–1} or W m^{–1} K^{–1}. The dimensions of k are [M L T^{–3} K^{–1}]

**Conduction Through a Composite Slab**

Case 1. Two slabs placed one on top of the other Suppose we have a composite slab made up of two different slabs of materials of thermal conductivities k_{1} and k_{2}, and cross-sectional areas A_{1} and A_{2} but of equal length L placed one on top of the other as shown in Fig. below

The ends of the composite slab are maintained at temperatures T_{1} and T_{2} (T_{1} > T_{2}). The rates of heat flow through each slab in the steady state are

The cross-sectional area of the composite slab is (A_{1} + A_{2}) but its length is L. If k_{eq} is the equivalent thermal conductivity, then the rate of heat flow through the composite slab is

Case 2. Two slabs placed in contact one after the other Suppose we have two slabs of different lengths L_{1} and L_{2} but of the same cross-sectional area A placed in contact as shown in Fig. below –

The ends of the composite slab are maintained at temperatures T_{1} and T_{2} (T_{1} > T_{2}). Let T0 be the temperature of the junction. In the steady state, Q_{1}/ t = Q_{2}/ t , i.e.

The rate of flow of heat through the composite slab is

The length of the composite slab is (L1 + L2) but its cross-sectional area is A. If k_{eq} is the equivalent thermal conductivity of the composite slab,

**Thermal Resistance**

Finding the equivalent thermal conductivity of a composite slab becomes much easier if we use the concept of thermal resistance. Just as electrical resistance is defined as

thermal resistance is defined as

(a) If two slabs are joined in series as shown in Fig. above, the equivalent thermal resistance of the composite slab is

(b) If the two slabs are joined in parallel as shown in Fig. above, then

All bodies emit heat from their surfaces at all temperatures. The heat radiated by a body is called radiant heat or thermal radiation. Thermal radiations are electromagnetic waves which travel in space with a velocity equal to that of light.

**(1) Black Body**

A perfect black body is one which completely absorbs radiations of every wavelength incident on it. A good absorber of radiations is also a good emitter of radiations. Consequently, a black body, when heated to a suitably high temperature, will emit radiations of all wavelengths. Such radiation is called black body radiation.

**(2) Emissive Power**

The emissive power (e) of a body is the amount of heat energy emitted per second from a unit area of a radiating surface. The SI unit of e is Js^{–1} m^{–2} or Wm^{–2}.

(3) Absorptive Power The absorptive power of a body is the ratio of radiant energy absorbed by it to the total amount of radiant energy incident on it. It is denoted by a and is a fraction. Since, by definition, a black body completely absorbs all radiations, a = 1 for a black body.

(4) Emissivity (ε** **) of a body is defined as the ratio of its emissive power (e) to that of a black body (E),

ε = e/E

For a black body, ε = 1.

**(5) Kirchhoff’s Law**

At any given temperature and for radiations of the same wavelength, the ratio of the emissive power to the absorptive power is the same for all substances, i.e

e/a = constant

**(6) Stefan’s Law**

The total energy emitted per second by a unit area of a body is proportional to the fourth power of its absolute temperature, i.e.

where σ is a constant known as Stefan’s constant. Its value is

**(7) Newton’s Law of Cooling**

The rate of loss of heat by a body is directly proportional to the excess of its temperature over that of its surroundings, provided this temperature difference is small, i.e

where T is the temperature of the body and T_{0} that of the surroundings. If m is the mass of the body, s its specific heat and dT the change in temperature in time dt, then dQ = msdT. Therefore,

This expression gives the time taken by a body to cool from T1 to T2 when placed in a medium of temperature T0. An approximate formula is

As the temperature of a black body increases, the maximum intensity of emission shifts (or is displaced) towards shorter wavelengths. In other words,

λ_{m}T = b = constant

where λ_{m} is the wavelength at which maximum emission takes place at absolute temperature T. The value of constant b is

b = 2.89 x 10^{–3} mK (metre kelvin)

## Thermodynamics

**(1) Equation of State**

In the case of ideal gases, the equation of state is

PV = RT, for one mole

and PV = nRT, for n moles

where P, V and T are the pressure, volume and temperature

of the gas, respectively and R is the universal gas constant.

**(2) First Law of Thermodynamics**

When heat energy is supplied to a system, a part of this energy is used up in raising the temperature of the system (i.e. in increasing the internal energy of the system) and the rest is used up in doing external work against the surroundings. Thus, if △Q is the heat energy supplied to a gas and if DW is the work done by it, then from the law of conservation of energy, the increase △U in the internal energy of the gas must be equal to (△Q – △W) or

△Q = △U + △W

Here all quantities are measured in energy units (joule). This equation is the mathematical statement of the first law of thermodynamics which may be stated in words as ‘if energy is supplied to a system which is capable of doing work, then the quantity of heat energy absorbed by the system will be equal to the sum of the increase in the internal energy of the system and the external work done by it.’ It must be remembered that, according to the accepted convention, △Q is positive if heat is supplied to the system and negative if heat is taken out of it and that DW is positive when work is done by the system and negative when work is done on the system. △U is positive if the temperature of the system increases and negative if the temperature decreases.

**(3) Internal Energy**

Every system (solid, liquid or gas) possesses a certain amount of energy. This energy is called the internal energy and is usually denoted by the symbol U. The internal energy of solid, liquid or gas consists of two parts: (i) kinetic energy due to the motion (translational, rotational and vibrational) of the molecules, and (ii) potential energy due to the configuration (separation) of the molecules. If the intermolecular forces are extremely weak or absent, then the change in internal energy is given by

△U = nC_{v} △T

**(4) Isothermal and Adiabatic Processes**

If a system is perfectly conducting to the surroundings and the temperature remains constant throughout the process, the process is called isothermal. During an isothermal process, the temperature of the system remains constant, but it can absorb heat from or lose heat to the surroundings. The process has to be extremely slow to be isothermal. During an adiabatic process, the system is completely insulated from the surroundings. It can neither give heat to nor take heat from the surroundings. Sudden processes are adiabatic such as the bursting of a cycle tube, the compression stroke in an internal combustion engine, etc. For an isothermal process,

PV = constant fi P_{1} V_{1 }= P_{2} V_{2}

For an adiabatic process

(5) Isobaric and Isochoric Processes If the pressure of the gas is kept constant, the process is called isobaric, i.e. P = constant or △P = 0. In an isochoric process, the volume of the gas is kept constant, i.e. V = constant or △V = 0.

(6) Work Done in a Process The work done by a gas in expansion from volume V1 to volume V2 is given by

**(1) Work Done in an Isothermal Process**

When an ideal gas is allowed to expand isothermally (i.e. at constant temperature) work is done by it. For an isothermal process, the equation of state for n moles of ideal gas is

PV = nRT

where T is the constant absolute temperature and R is the universal gas constant. The value of R for all gases is R = 8.31 J K^{–1} mol^{–1}. From Eq. (1)

where V1 is the initial volume and V2, the final volume of gas. In terms of initial and final pressures, P1 and P2, Eq. (2) can be written as (since P1V1 = P2V2)

**(2) Work Done in Adiabatic Process**

When a gas undergoes an adiabatic change, the pressure-volume changes obey the relation

PV y = constant = C

where y = Cp/Cv is the ratio of specific heat of the gas at constant pressure to that at constant volume. From Eq. (3), we have

The work done is given by

since P1V1 = nRT1 and P2V2 = nRT2, T1 and T2 being the absolute temperatures before and after the adiabatic change.

**(3) Work Done in an Isobaric Process**

For an isobaric process, pressure P is constant. Therefore, the work done is given by

**(4) Work Done in an Isochoric Process**

In an isochoric process, volume V is constant, i.e. dV = 0. Hence, work done W = 0.

**(7) Indicator Diagram**

If the exact relationship between P and V is not known, the above expressions for work done cannot be used. In such cases, the graphical (indicator diagram) method is employed. Indicator diagram (or P – V diagram) is a graph in which pressure (P) is plotted on the y-axis and volume (V) on the x-axis. Figure 8.7 shows indicator diagrams for expansion, compression and for a closed cyclic process. The initial state (P1, V1) is represented by point A and the final state

(P2, V2) by point B. The intermediate states are represented by points between A and B on the curve AB.

Work done in process A → B = area under AB in Fig. 8.7(a).

Work done in process B → A = area under AB in Fig. 8.7(b).

Work done in the cyclic process A → B → A is equal to the area enclosed by the loop.

**(8) Efficiency of an Ideal Heat Engine**

The efficiency (η) of an ideal reversible heat engine is given by

η = 1 – T_{2}/T_{1}

where T_{1} = absolute temperature of the source which supplies heat and

T_{2} = absolute temperature of the sink which takes in the part of heat not converted into useful work,

T_{1} is always greater than T_{2}. If T_{1} = T_{2}; η = 0 implying that an engine working under isothermal conditions can produce no useful work. Complete conversion of heat into work (i.e. η = 1) is possible only if T_{2} = 0, i.e. the sink is at absolute zero, which is unattainable.

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