### The Misconception of Heat Unravelled

Most books define heat as if it were a form of energy. Heat transfer is dealt as the transfer of heat from one control volume to another control volume. Nothing can be farther from the truth. Not only does this kind of definition or representation give a false intuitive idea to the newbie, but it also causes confusion since the person would notice that something doesn't seem to fit in when he/she looks at the thermodynamic equations.

Before I had my College Physics classes, I was a victim to this admittedly. I thought heat was transferred from one point to another and whose abundance was dependent on temperature. The first time I've read of a warning that this perspective was wrong was when I was doing advanced readings a few months before the start of classes (that is, the classes which include College Physics - somehow, reading the materials over the summer seems to help make things easier to understand). It said that a popular convention, and an extremely erroneous one, was to term the amount of energy transfer as heat or heat transfer. It also said that this was so ingrained in the academe that it was a bit futile to combat this malady. In the end, the textbook just asked its reader to be patient with any misuse of the term and keep in mind the proper way of using it. And so, for 5 years, I held my silence until now, when I started writing here in Schematic Analysis.

Anyway, now you know that this isn't something I just made up. Heat transfer really is a misleading term. Bottom line is, heat transfer refers to the amount of energy transferred from one system to another, a MEASURE of the RATE of energy transfer, NOT A FORM of energy. It is like referring to velocity as a form of matter. Considering the term heat alone, it is simply the feeling of a body towards another body that is at a higher temperature. So, instead of saying "Calculate the amount of heat transferred.", we say "Calculate the amount of energy transferred by heat.". When we hear a weather broadcast saying that tomorrow will be a day of scorching heat, we know that the heat there isn't concrete and definitely isn't scorching, but an energy transfer.

What will happen if we disregard the above statements and look at the equations?

Let us see:

Q=m*c*(Change in temperature)

Where:
Q - heat developed
m - mass of the system
c - specific heat (usually at room temperature)

It looks like I can create heat energy out of nothing at all by simply putting into contact two systems that are at different temperatures (they will try to reach thermal equilibrium). But this is a complete violation of the Law of Conservation of Energy - energy can not be created or destroyed. How can this be?

Now back to the correct interpretation, heat developed interpreted as a rate of energy transfer. Two systems at different temperatures are placed in contact. The system at lower temperature tends to have its temperature increased (because the particles are more excited by the higher temperature of the other system). This increase in temperature is a rate of energy trasfer because energy was transferred from the hotter system to the colder system, thus "heating" it. We can say that this rate of energy transfer is proportional to the amount of temperature increase or decrease of the system. Therefore, the equation makes sense. Of course, this increase and decrease is dependent on how much of the system there is (mass) and what material the system is composed of (specific heat constant).

### Calculator Techniques for the Casio FX-991ES and FX-991EX Unraveled

In solving engineering problems, one may not have the luxury of time. Most situations demand immediate results. The price of falling behind schedule is costly and demeaning to one's reputation. Therefore, every bit of precaution must be taken to expedite calculations. The following introduces methods to tackle these problems speedily using a Casio calculator FX-991ES and FX-991EX.

►For algebraic problems where you need to find the exact value of a dependent or independent variable, just use the CALC or [ES] Mode 5 functions or [EX] MENU A functions.

►For definite differentiation and integration problems, simply use the d/dx and integral operators in the COMP mode.

►For models that follow the differential equation: dP/dx=kt and models that follow a geometric function(i.e. A*B^x).

[ES]
-Simply go to Mode 3 (STAT) (5)      e^x
-For geometric functions Mode 3 (STAT) 6 A*B^x
-(Why? Because the solution to the D.E. dP/dx=kt is an exponential function e^x.
When we know the boundary con…

### Common Difficulties and Mishaps in 6.004 Computation Structures (by MITx)

Updated:
May 6, 2018
VLSI Project: The Beta Layout [help needed]Current Tasks: ►Complete 32-bit ALU layout [unpipelined] in a 3-metal-layer C5 process. ►Extend Excel VBA macro to generate code for sequential instructions (machine language to actual electrical signals).
Current Obstacles/Unresolved Decisions:
►Use of complementary CMOS or pass transistor logic (do both? time expensive, will depend on sched.
►Adder selection: Brent-Kung; Kogge Stone; Ladner Fischer (brent takes up most space but seems to be fastest, consider fan-out) [do all? time expensive, will depend on sched.)
►layout requirements and DRC errors