Here are the

**(1)** Temperature *t*_{C} in Celsius scale can be converted to temperature *t*_{F} in Fahrenheit scale using the relation

*t*_{C}/100 = (*t*_{F }– 32)_{ }/180

Remember that the ice point and steam point of water are 32º F and 212º F respectively in the Fahrenheit scale. These *temperature difference* (*∆t*_{F}) of 180º in the Fahrenheit scale is equal to a *temperature difference* (*∆t*_{C}) of 100º in the Celsius scale. The significance of the numbers 100, 32 and 180 will be clear to you now.

**(2) **Temperature *t*_{C} in Celsius scale can be converted to temperature *t*_{K} in Kelvin scale (absolute scale) using the relation

*t*_{C}/100 = (*t*_{K }– 273.15)_{ }/100

so that *t*_{C} = *t*_{K }– 273.15 or, *t*_{K} = *t*_{C }+ 273.15.

Often the above relation is written as** t_{K} = t_{C }+ 273 **very nearly.

The above equation follows from the fact that the ice point and steam point of water are 273.15 K and 373.15 K respectively in the Kelvin scale.

A *temperature difference* of 1º in the Kelvin scale is equal to a* temperature difference* of 1º in the Celsius scale.

**(3) **The *increase* in length (*∆**ℓ*) of a solid on raising its *∆T* is given by

*∆**ℓ*** = α ℓ_{o }**

*∆T**where*

*α*is the coefficient of

**linear expansion (**

*linear expansivity*) and

*ℓ*

_{o}is the original length.

**(4) **The *increase* in area (*∆**A*) of a solid on raising its *∆T* is given by

*∆A*** = β A_{o }**

*∆T**where*

*β*is the coefficient of

**area expansion (**

*area expansivity*) and

*A*

_{o}is the original area.

In the case of isotropic homogeneous solids *β = *2*α*

**(5) **The *increase* in volume (*∆**V*) of a solid on raising its *∆T* is given by

*∆V*** = γV_{o }**

*∆T**where*

*γ*is the coefficient of

**volume expansion (**

*volume expansivity*) and

*V*

_{o}is the original volume.

In the case of isotropic homogeneous solids *γ = *3*α*

**(6) Specific heat capacity ( specific heat) C** of a substance is the quantity of heat absorbed or rejected by 1 kg of the substance to change the

Therefore, the heat involved in changing the *m *kg of a substance through *∆T* K (or *∆T* º C) is *mC* *∆T* where *C* is the specific heat of the substance.

**(7) Molar specific heat **of a substance is the quantity of heat absorbed or rejected by 1 mole of the substance to change the

You know that in the case of gases there are two specific heats viz., ** specific heat at constant volume** and

**. Similarly there are two molar specific heats viz., molar specific heat at constant volume and molar specific heat at constant pressure. In this context you will find a useful post here.**

*specific heat at constant pressure***(8) **Heat transfer takes place by three processes viz., conduction, convection and Radiation.

The quantity of heat *Q* conducted in a time *t *through a uniform rod of length *L *and area of cross section *A *when the ends of the rod are maintained at a temperature difference *∆T* is given by

*Q* = *KA* (*∆T/L***) t** where

*K*is the

**of the material of the rod**

*thermal conductivity*[*The temperature gradient*** ***∆T/L* is often written as *∆T/∆x*, replacing *L* by *∆x*, which will be more appropriate in the case of heat conduction through slabs of thickness *∆x*].

The rate of transfer of heat by conduction (time rate), *H* is given by

*H** = Q/t = KA*

*∆T/L* According to **Stefan’s law** the energy (*E*) *radiated* per second per unit surface area of a perfectly black body is directly proportional to the 4^{th} power of the absolute temperature of the body:

*E* α *T*^{4}

Or, *E = **σT*^{4} where *σ* is Stefan’s constant.

So if the temperature is doubled, the energy *radiated *from the body will become 16 times.

*dQ/dt *α (*T*_{2} – *T*_{1}) where* dQ* is the heat lost in a time *dt* when the temperatures of the body and the surroundings are respectively *T*_{2} and *T*_{1}.

Note that the loss of heat mentioned in *T*_{2} – *T*_{1}) should be small. * *

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