Temperature and Heat for AP Physics B- Equations to be Remembered

Here are the essential things you need to remember for solving multiple choice questions involving heat and temperature:

(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 temperatures in the Celsius scale are 0º C and 100º C respectively so that a 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 temperature by ∆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 temperature by ∆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 = 2α

(5) The increase in volume (∆V) of a solid on raising its temperature by ∆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 = 3α

(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 temperature by 1 K.

Therefore, the heat involved in changing the temperature of 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 temperature by 1 K.

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

(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 thermal conductivity of the material of the rod

[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 T⁴

Or, E = σT⁴ where σ is Stefan’s constant.

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

Newton’s law of cooling says that the rate of cooling (rate of loss of heat) of a body is directly proportional to the excess of temperature of the body over the surroundings:

dQ/dt α (T₂ – T₁) where dQ is the heat lost in a time dt when the temperatures of the body and the surroundings are respectively T₂ and T₁.

Note that the loss of heat mentioned in Newton’s law of cooling is due to all the three mechanisms viz., conduction, convection and radiation. Further, the excess temperature (T₂ – T₁) should be small.

In the next post we will discuss questions in this section. Meanwhile find a useful post here.