**Where are we going with this?**This page will assist in developing the ability to

*p*

*erform calculations involving heat flow, temperature changes, and phase changes by using known values of specific heat, phase change constants, or both.*

**Thermal Energy (Heat) Change and Transfer**

*Again with the two names? Come on!*

Thermal energy… heat… *Q*… whatever…

**In this article we are going with "heat" and " Q" as we talk about the amount of thermal energy that is… doing something. **

So, we discussed in a previous article that the amount of heat related to some temperature change of some substance can be found easily:

Where *Q* is heat change, where *m* is mass, *c *is specific heat, and *∆T* is change in temperature,

Q = mc∆T

**It's fair to think of it this way…**

Qis heat (thermal energy)mis how much stuff you have measured in masscis what kind of stuff you have, specifically the physical property of specific heat∆Tis how much did the temperature of the stuff change

Seems like a pretty good time for an example problem. But first… a little nit-picky thing… that

*c*…cis a constant that relates mass, temperature change and type of substance to heat. So, it has to have units that match what you are working with. Standard science units for heat is joules, and for mass kilograms.So, for example, c for water is 4128 joules per kilogram per degree C.That means if one kilogram of water cools off by 1°C, it gives off 4128 joules. What's the point?The point is this:depending on what mass and energy units you are using, you need to make sure you are using the correct value for specific heat.

Okay… examples…

________________________

A cast iron brake rotor of a moderately interesting vehicle, having a mass of 9.5 kg, undergoes an increase in temperature from 25°C to 35°C. How much heat energy was absorbed? The specific heat for cast iron is 449 j/kg • C° .Step 1: Find the temperature change.WhereTfrepresents final temperature andTirepresents initial temperature, then∆T = Tf - Ti∆T = 35 - 25∆T = 10Step 2: Use ∆T to find QQ = mc∆TQ = 9.5 • 449 • 10Q = 42,655 J

________________________

*Easy! Now What? There must be some way to make it harder!*

Finding something besides Q…

Sometimes, you will be given Q and asked to find one of the other numbers. In such a case, you just use the same equation and solve.

________________________

A sample of metal gives off 50,000 J of heat. If the temperature change was 10°C and the specific heat was 250 J/kg • °C, what was the mass?

Given any two of the three things, the algebra is the same, except you solve for a different number.

Q = mc∆T

50,000 = m • 250 • 10

50,000 = m • 2500

50,000 / 2500 = m

20 kg = m

### Working with Heat Exchange

The calculations become longer (but equally easy) when dealing with cases where one substance exchanges heat with another substance. Now, there's more parts, but still all easy parts.

Usually, you'll be given a lot of info about one of the substances, then asked to find one part of the info for the other substance. Something like…

A 4 kg chunk of lead with an initial temperature of 300°C is placed into water having an initial temperature of 59.375°C and and cooled off. The final temperature of the mixture is 75°C.What was the mass of the water?The specific heats:water = 4128lead = 129

*You must be joking!*

It's not too bad.

*Because…* HEAT LOST = HEAT GAINED!

So, the lead cooled down, giving off heat. The water heated up, gaining heat.

If we find the heat lost by the lead, that's a number. Then, we use

*THAT*number as the heat gained by the water. Then, we plug in and solve. Actually, it's just a lot of steps, but*not*hard ones.Let's do it!

________________________

A 4 kg chunk of lead with an initial temperature of 300°C is placed into water having an initial temperature of 59.375°C and and cooled off. The final temperature of the mixture is 75°C.What was the mass of the water?

We need to look up the specific heats:

water = 4128lead = 129

PART 1: Heat Lost by the LeadStep 1: Find the temperature change of the lead.WhereTfrepresents final temperature andTirepresents initial temperature, then∆T = | Tf - Ti|∆T = 300 - 75∆T = 225Step 2: Use ∆T to find QQ = mc∆TQ = 4 • 129 • 225Q = 116,100 JHeat Lost = 116.100 JPART 2: Heat Gained by the Water and finding what is not givenStep 1: Heat GainedHeat Lost = Heat Gained116,100 J = Heat GainedStep 2: Find the temperature change of the water.WhereTfrepresents final temperature andTirepresents initial temperature, then∆T = | Tf - Ti |∆T = | 75 - 59.375 |∆T = 15.625Step 3: Find what is missing (mass, m, in this case)Q = mc∆T116,100 = m • 4128 • 15.625116,100 / (4128 • 15.625) = mcalculator… make sure you do the parentheses first!1.8 kg = mThere was 1.8 kg or water.

________________________

All in all, the whole heat change, heat exchange process is pretty easy. There ARE lots of steps involved,

*but*each step is easy.
## No comments:

## Post a Comment