Heat Transfer and Calorimetry IIT JEE Study Material

Heat is a form of energy that is transferred between a system and its surrounding as a result of temperature difference. The expansion due to an increase in the temperature is known as thermal expansion. There are three types of thermal expansion, namely Linear, Superficial, and Volume expansion.

Types of Thermal Expansion:

1. Linear Expansion:

When there is any change in the length of a body due to heating then the expansion is called longitudinal or linear expansion.

Coefficient of Expansion: Δ1 = Change in length and Δt = Change in Temperature

α  =  limΔt    0    1l0    ΔlΔt    and    Δl  =  l0  α  Δt\mathbf{\alpha \;=\;\lim_{\Delta t\; \rightarrow \;0}\;\;\frac{1}{l_{0}}\;\;\frac{\Delta l}{\Delta t}\;\;and\;\;\Delta l\;=\;l_{0}\;\alpha \;\Delta t}

2. Superficial Expansion:

When there is any change in the area of a body due to heating then the expansion is called axial or superficial expansion.

Coefficient of Expansion: ΔA = Change in Area and Δt =  Change in Temperature

β  =  limΔt    0    1A0    ΔAΔt    and    ΔA  =  A0  β  Δt\mathbf{\beta \;=\;\lim_{\Delta t\; \rightarrow \;0}\;\;\frac{1}{A_{0}}\;\;\frac{\Delta A}{\Delta t}\;\;and\;\;\Delta A\;=\;A_{0}\;\beta \;\Delta t}

3. Volumetric Expansion:

When there is any change in the volume of a body due to heating then the expansion is called volumetric or cubic expansion.

Coefficient of Expansion: ΔV = Change in volume and Δt = Change in Temperature

γ  =  limΔt    0    1V0    ΔVΔt    and    ΔV  =  V0  γ  Δt\mathbf{\gamma \;=\;\lim_{\Delta t\; \rightarrow \;0}\;\;\frac{1}{V_{0}}\;\;\frac{\Delta V}{\Delta t}\;\;and\;\;\Delta V\;=\;V_{0}\;\gamma\;\Delta t}

If α1, α2, and α3 are coefficient of linear expansion in X, Y  and Z directions, then,

Case 1: For Isotropic Solids

α = α1 = α2 = α3,  β = 2α  γ = 3α

Case 2: For Anisotropic Solids

β = α1 + α2  and γ = α1 + α2+ α3

Thermal Stress:

If temperature of a rod of length l0­ clamped between two fixed walls separated by same distance l0 is changed by amount Δt then,

Case 1: α is constant

Stress = F / A and Strain = Δ1 / 1o

Therefore, Young’s Modulus = FAΔll0  =  F  l0A  Δl  =  FA  α  Δt\mathbf{\frac{\frac{F}{A}}{\frac{\Delta l}{l_{0}}}\;=\;\frac{F\;l_{0}}{A\;\Delta l}\;=\;\frac{F}{A\;\alpha \;\Delta t}}

i.e. F  =  Y  A  α  Δt\mathbf{F\;=\;Y\;A\;\alpha \;\Delta t}

Case 2: α is not constant

A. If α varies with distance [α = ax + b]

Total thermal expansion = 01    (ax  +  b)  dx  Δt\mathbf{\int_{0}^{1}\;\;\left ( ax\;+\;b \right )\;dx\;\Delta t}

B. If  α varies with temperature [α = f (T)]

Δl  =  T1T2  α  l0  dT\mathbf{\Delta l\;=\;\int_{T_{1}}^{T_{2}}\;\alpha \;l_{0}\;dT}

Variation in Density: Density decreases with an increase of temperature because of the increase in volume and vice-versa i.e.

Density d = d01  +  γ  Δt\mathbf{\frac{d_{0}}{1\;+\;\gamma \;\Delta t}}

Special Case: The Density water is maximum at 4 °C. Its density increases from 0 °C to 4 °C (g is positive). From 4 °C to higher temperatures g is positive.

Modes of Heat Transfer:

While conduction is the transfer of heat energy by direct contact, convection is the movement of heat by actual motion of matter; radiation is the transfer of energy with the help of electromagnetic waves.

Heat Transfer Conduction Convection Radiation
Definition Heat is transferred between objects by direct contact Energy transition (heat transfer) occurs within the fluid Heat is the transmission is done without any physical contact between objects
Representation How heat travels between objects in direct contact How heat passes through fluids How heat flows through empty spaces
Occurrence Occurs in solids through molecular collisions. Occurs in fluids by the actual flow of matter. Occurs at a distance and does not heat the intervening substance.
Causes temperature difference density difference Occurs from all objects, at a temperature greater than 0 Kelvin
Speed Slow Slow Fast
Law of reflection and refraction Does not follow Does not follow Follow
Transfer of heat Uses heated solid substance Uses intermediate substance Uses electromagnetic waves

Differential form of Ohm’s law:

dQdT  =  K  A  dTdx\mathbf{\frac{dQ}{dT}\;=\;K\;A\;\frac{dT}{dx}}

Where, dT / dX  = Temperature Gradient

Stefan-Boltzmann law:

It states that the net radiant heat energy emitted from an object is proportional to the 4th power of its absolute temperature.

E  =  σ  A  T4    J  sec1  m2\mathbf{E\;=\;\sigma \;A\;T^{4}}\;\;J\; sec^{-1} \;m^{-2}

Radiation Power:

dQdT  =  σ  A  T4    watt\mathbf{\frac{dQ}{dT}\;=\;\sigma \;A\;T^{4}\;\;watt}

If Ts is the surrounding temperature:   [Black Body]

dQdT  =  σ  A  (T4    Ts4)\mathbf{\frac{dQ}{dT}\;=\;\sigma \;A\;(T^{4}\;-\;T_{s}^{4})}

Emissive Power or Emissivity e = Heat from given body / Heat from a black body

Newton’s Law of cooling:

It states that the rate of change of the temperature (T) of an object is proportional to the difference between its own temperature and the temperature of its surroundings.

T(t) = Ts + (To – Ts ) e – k t

Where,

Ts  = Surrounding Temperature

To = Initial temperature of the body

T (t) = Temperature of the body at time t

k = Cooling constant

Wien’s black body radiation:

At any temperature T greater than 0 Kelvin the body emits energy radiations of all wavelengths. According to the Wien’s displacement law, if the wavelength (λ) corresponding to the maximum energy is λmax then,

λmax T = b

Where,

T = Temperature of the body,

b = Wien’s Constant.

Calorimetry

It is the measurement of changes in the state variables of an object for the purpose of deriving the transfer of heat associated with changes in its state either due to phase transitions, physical changes, or chemical reactions, under specified constraints. A calorimeter is a device using which Calorimetry is performed.

The amount of heat required to raise the temperature of 1 gram of water from 14.5° C to 15.5°C at Standard Temperature and Pressure (STP) is 1 calorie.

Q  =  m  T1T2    C  dt  =  m  C  ΔT\mathbf{Q\;=\;m\;\int_{T_{1}}^{T_{2}}\;\;C\;dt\;=\;m\;C\;\Delta T}

Heat transfer in phase change:

Q = mL

Where,

L = latent heat of substance in Cal gm-1 0C-1 or in Kcal kg-1 0C-1

Lsteam = 540 cal/ gm

Lice = 80 cal/ gm for ice

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