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# Heat

## The effects of heat

Heat produces two major effects on the matter which receives or gives up heat : - a variation of temperature, or - a variation of physical state (solid, liquid, gaseous) of the matter

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## The transfer of heat

### Conduction

Conduction" is the transfer of heat in solid matter. Metals have a better thermal conductivity than non-metals. (Example: The metal spoon in the hot coffee)

"Thermal radiation" is the transfer of heat by infrared rays (Part of the electromagnetic spectrum corresponding to radiation of a greater wavelength than visible light and smaller than microwave). All bodies, provided their temperature Kelvin is not zero, emit thermal radiation. (Example: An oven)

### Convection

Heat causes migration of molecules in a liquid, generally from bottom to top because warm liquids are less dense than cold ones. (Example: Central warm water heating)

## The received amount of heat

You need two times the quantity of heat to have a twofold rise in temperature: The amount of heat is proportional to the elevation of temperature. You need two times the amount of heat to produce the same rise of temperature of a double mass: The amount of heat is proportional to the mass. Heating different matters, a different result is obtained with the same amount of heat: The amount of heat depends on the matter.

$\Delta\; Q$ $=$ $c\;m\;\Delta\; T$ $\Delta\; Q$ : received heat $\Delta\; T$ : variation of temperature $m$ : mass $c$ : specific heat capacity (constant) depending on the matter.

## The calorie

The calorie is the amount of heat which rises the temperature of $1 \;g$ water by $1^oC$.

Introducing this definition in the fundamental formula, it follows : $\Delta\; Q$ $=$ $c\;m\;\Delta\; T$ $1\;cal$ $=$ $c\dot 1\;g\cdot 1\;K$ $c$ $=$ $\frac{1\;cal }{1g\;1K}$ $c$ $=$ $1 \frac{cal}{g \cdot K}$

$c_{H_2O}$ $=$ $1 \;\frac{cal}{g \cdot K}$

## Exercise

A lump of sugar has a mass of $3\; g$. The combustion of $1\; g$ of sugar liberates $4\; kcal$. How many lumps must theoretically be burnt to heat $600\; g$ coffee (same $c$ than water) from $20^oC$ to $100^oC$? Answer: $\Delta Q$ $=$ $c\cdot m\Delta T$ $\Delta Q$ $=$ $1\cdot 600 \cdot (100$ $-$ $20)$ $=$ $48000\; cal$ $=$ $48\; kcal$ $\frac{48}{4}$ $=$ $12\;g$ of sugar must be burnt, that means 4 lumps!