Watt & heat calculation this is my first post in this forum. Unfortunately, I'm not very familiar with physics and I hope I'm in the right place here. I asked myself how much energy I need to heat something.
I've found that by wattage you can figure out heat in Celcius.
So I researched the net and found that 1 watt is about 1.16194 C˙.
Then if a processor, as installed in my Mac, draws 65W, the chip would have to warm up to 55.94 C˙. But that's not true, because I measure temperatures of 105 degrees.
How exactly or what exactly am I doing wrong? For example, how can I find out how hot the surface of a light bulb draws 100W or in the case of a processor.
 A: 
So I researched the net and found that 1 watt is about 1.16194 C

Technically not correct. The temperature difference is
$$\Delta T = T_2 - T_1 = Q \cdot R$$
where $T$ is the temperature in $^\circ\text{C}$, $Q$ is the generated heat power in $\text{W}$, and $R$ is the thermal resistance in $^\circ\text{C/W}$. In your case, temperature of the processor would be
$$\boxed{T_2 = T_1 + Q \cdot R}$$
where $T_1$ is the ambient temperature, $Q$ is heat generated by the processor, and $R$ is thermal resistance between the processor and the ambient. Note that ambient is actually the fluid that cools down the processor. For simple air cooling the ambient temperature is approximately equal to the room temperature.
From the data you have given us, the thermal resistance is $R = 1.16194 \text{ }^\circ\text{C/W}$ which means that for every $1 \text{ W}$ of generated heat power the temperature rises by $1.16194 \text{ }^\circ\text{C}$. For $Q = 65 \text{ W}$ of heat generated by the processor, the processor temperature would be
$$T_2 = T_1 + 75.5 \text{ }^\circ\text{C}$$
If the ambient temperature is about $T_1 = 25 \text{ }^\circ\text{C}$ the processor temperature is about $100 \text{ }^\circ\text{C}$, which is quite close to the measured processor temperature.
It must be noted that this model is overly simplified and you should not expect super-accurate results. In everyday engineering, anything within 5-10 $^\circ$C is considered to be good enough!
A: Sorry for You, but "I've found that by wattage you can figure out heat in Celcius." is not true. Degrees give a state of a body, not its energy. 1 Watt is a measure of energy per time, so with about 4 watt you could change the temperature of   1 g of water by 1°C in 1 second so from 20°C to 21°C or from 7°C to 8°C
if your mac draws 65W   it depends very much how long You use it, how your processor is cooled, how warm the rest of its surrounding is. You will never be able to calculate the temperature by the watts.
A: 
I've found that by wattage you can figure out heat in Celcius.
So I researched the net and found that 1 watt is about 1.16194 C˙.
Then if a processor, as installed in my Mac, draws 65W, the chip would
have to warm up to 55.94 C˙. But that's not true, because I measure
temperatures of 105 degrees.

Unfortunately this completely wrong.
But it does take 'power' to increase the temperature of an object.
Firstly let's do some defining.

*

*power:

In physics power $P$ is defined as the ratio of energy to time. Assuming that in our case $P$ is constant in time:
$$P=\frac{\Delta E}{\Delta t}$$
where $\Delta E$ the amount of energy per time interval $\Delta t$.
E.g. a light bulb of $60\,\mathrm{W}$ uses (and radiates) $60\,\mathrm{J}$ ('Joule') per second of energy.

*

*temperature increase due to energy input:

Energy input to the object and temperature increase of the object are related as follows:
$$\Delta E=mc_p\Delta T\tag{1}$$
where $m$ is the mass of the object, $c_p$ the object's specific heat capacity and $\Delta T=T_2-T_1$ the temperature increase the object will experience.
$c_p$ (in $\mathrm{J kg^{-1}K^{-1}}$) is the increase in temperature if we add $1\,\mathrm{Joule}$ to $1\,\mathrm{kg}$ of object, in $\mathrm{Kelvin}$ (but here we can perfectly use $^{\circ}\mathrm{C}$ - degrees centigrade)

So $(1)$ tells you what happens we we put a discrete amount of heat energy $\Delta T$ into an object but what happens is we put power, i.e. a continuous flow of energy into the same object:
$$\frac{\Delta E}{\Delta t}=mc_p\frac{\Delta T}{\Delta T}\tag{2}$$
This would seem to indicate that if we just keep up the power we could reach any temperature, right?
As it happens, also with your processor, most heated object do not only receive energy, they also lose energy.
This energy loss may be intended (cooled car engines e.g.) or circumstantial (your processor loses energy in a variety of way)
This loss of heat prevents the temperature of the object from rising and rising, so is actually much desired.
Some final, 'steady state' temperature will be achieved when:
$$P_{in}=P_{out}$$
But it can't be found merely from input power.
