# Thermal

Written by Jerry Ratzlaff on . Posted in Thermodynamics

## Thermal Conductivity

Thermal conductivity is the ability to transfer heat within a material without any motion of the material.  Depending on the material, the transfer rate will vary.  The lower the conductivity, the slower the transfer.  The higher the conductivity, the faster the transfer.

### Thermal Conductivity formula

$$\large{ k = \frac{Ql}{A \Delta T} }$$

Where:

$$\large{ k }$$ or $$\large{ \lambda }$$   (Greek symbol lambda) = thermal conductivity

$$\large{ \alpha }$$   (Greek symbol alpha) = thermal diffusity (heat transfer rate)

$$\large{ A }$$ = area of the object

$$\large{ l }$$ = length or thickness of material

$$\large{ \rho }$$   (Greek symbol rho) = density

$$\large{ Q }$$ = amount of heat transfer through a material

$$\large{ \Delta T }$$ = temperature differential

## Thermal Conductivity Constant

### Thermal Conductivity Constant formula

$$\large{ k_t = \frac {\dot {Q}_t l} {\Delta T} }$$

Where:

$$\large{ k_t }$$ = thermal conductivity constant

$$\large{ l }$$ = length

$$\large{ \dot {Q}_t }$$ = heat transfer rate

$$\large{ \Delta T }$$ = temperature differential

Solve for:

$$\large{ l = k_t \frac {\Delta T} {\dot {Q}_t} }$$

## Thermal Diffusivity

Thermal diffusivity ( $$a$$ ) is a measure of the transient thermal reaction of a material to a change in temperature.

### Thermal Diffusivity formula

$$\large{ \alpha = \frac {k_t} {p Q } }$$

Where:

$$\large{ \alpha }$$   (Greek symbol alpha) = thermal diffusity (heat transfer rate)

$$\large{ k_t }$$ = thermal conductivity constant

$$\large{ p }$$ = density

$$\large{ Q }$$ = specific heat capacity

Solve for:

$$\large{ k = \alpha p Q }$$

$$\large{ p = \frac {k} {\alpha Q} }$$

$$\large{ Q = \frac {k} {\alpha p} }$$

## Thermal Energy

Thermal energy ( $$Q$$ or $$TE$$ ) (also called heat energy and heat transfer) is the exertion of power that is created by heat, or the increase in temperature.

### Thermal Energy formula

$$\large{ Q = mc \Delta T }$$

Where:

$$\large{ Q }$$ or $$\large{ TE }$$ = thermal energy

$$\large{ c }$$ = specific heat

$$\large{ m }$$ = mass

$$\large{ \Delta T }$$ = temperature differential

Solve for:

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

$$\large{ \Delta T = \frac {Q}{mc} }$$