Physics Maths Problem

[Queztacotl]

Here is a problem that I have absolutely no idea about:

 

You make a cup of coffee out of 300g of coffee at 100 Degrees Celsius and 30g of pure ethyl alcohol at 20 Degrees Celsius. One Joule is enough energy to produce a change of 1 Degree Celsius in .42g of ethyl alcohol ( i.e ,alcohol is easier to heat then water). What temperature is the final mixture?

 

Adding up all the energy after mixing has to give the same result as the total before mixing . We let the subscript i stand for the initial situation , before mixing and f for the final situation , and use subscripts c for the coffee and a for the alcohol. In this notation, we have :

total initial energy = total final energy

Eci + Eai = Ecf + Eaf

 

We assume coffee has the same heat-carrying properties as water. Our information about the heat-carrying properties of the two substances is stated in terms of the change in energy required for a certain change in temperature , so we re-arrange the equation to express everything in terms of energy differences:

 

Eaf - Eai = Eci - Ecf

 

Using the given ratios of temperaturechange to energy change we have :

 

Eci - Ecf = ( Tci - Tcf ) ( Mc ) / ( 0.24g)

Eaf - Eai = ( Taf - Tai) ( Ma ) / ( 0.42g)

 

Setting these two quantities to be equal, we have :

 

( Taf - Tai) ( Ma) / ( 0.42g ) = (Tci - Tcf ) ( Mc) / (0.24g)

 

In the final mixture the two substances must be at the same temperature, so we can use a single symbol Tf= Tcf = Taf, for the two quantities previously represented by two different symbols,

 

( Tf - Tai) ( Ma) / (0.42g) = (Tci- Tf) ( Mc) / (0.24g)

 

Solving for Tf gives :

 

Tf = Tci with Mc /.24 + Tai with Ma / .42

Mc/ 0.24 + Ma / 0.42

 

= 96 Degrees Celsius

 

I dont understand any part fully and don't know where T come from or any mathematical part fully because I havent done anything like this before. Only reply to this if you know the full understanding of the question and how the formula and all is used!

[CraigD]

Some of you equations don’t follow from one another, but in the end, your result agrees with mine.

 

You may find it helpful to think of the problem in terms of adding energy, using concrete examples.

 

Imagine you started with the 300 g of water at 10 degrees, and added energy until it’s at 100.

 

Change in energy = mass * specific heat * change in temperature.

Where: the specific heat of water is ~ 4.2 J/(g*degree); of alcohol, ~2.4

 

So, the energy added = 300*4.2*(100-10) = 113400

 

Now imagine you start with 30 grams of alcohol at 10 degrees, and add energy until it’s at 20 degrees.

 

Energy added =30*2.4*(20-10) = 720

 

Now imagine we add both amounts of energy to a mixture of the 2 liquids at 10 degrees.

 

Change in temperature = Change in energy / (mass * specific heat)

Where the mixture’s mass * specific heat = mass1*specific heat1 + mass2*specific heat2

 

= (113400+720) / (300*4.2+30*2.4) = ~ 86

 

So, the heat of the mixture would be increased from 10 to about 96 degrees.

 

It doesn’t matter how the energy is added to the mixture. Adding it to the water and alchohol separately, then mixing them, is the same as mixing them, then adding to the mixture.

 

This sort of problem can become complicated if the changes in temperature involve changes in the physical phase of the masses – solid – liquid – gas – since specific heat is different for the different phases. Also there’s a slight change in the specific heat of water with change in temperature. But everything is in liquid phase, and for the precision we’re using, the specific heat changes can be ignored.

[Queztacotl]

Imagine you started with the 300 g of water at 10 degrees, and added energy until it’s at 100.

 

Change in energy = mass * specific heat * change in temperature.

Where: the specific heat of water is ~ 4.2 J/(g*degree); of alcohol, ~2.4

 

So, the energy added = 300*4.2*(100-10) = 113400

 

Now imagine you start with 30 grams of alcohol at 10 degrees, and add energy until it’s at 20 degrees.

 

Energy added =30*2.4*(20-10) = 720

 

Now imagine we add both amounts of energy to a mixture of the 2 liquids at 10 degrees.

 

Change in temperature = Change in energy / (mass * specific heat)

.

 

So when your changing it in energy, you multiply mass * specific heat * Change in temperature always?

 

I don't understand what ~ 4.2 J/(g*degree); of alcohol, ~2.4 means. Whats the g/degree for and how does this part fit in with the whole question?

 

I haven't done this type or near as any type of a question like this or similar to this before so be pretty basic to lead up to the points your making!

[CraigD]

So when your changing it in energy, you multiply mass * specific heat * Change in temperature always?

Yes. So you need specific heat tables for the various compounds in which you're interested. I used this table, which also has a good explanation of specific heat.

I don't understand what ~ 4.2 J/(g*degree); of alcohol, ~2.4 means. Whats the g/degree for and how does this part fit in with the whole question?

Specific heat is measured in units of energy-per-mass-temperature - Joules/kilograms*degrees Celsius in the International System (SI), British Thermal Units/Pound*degrees Fahrenheit in the English system. To make the numbers smaller, I put them in Joule/gram*degree by dividing SI data from a table by 1000. I abbreviated Joules and grams by their 1st letter. The punctuation characters I use mean the following:

~ about, approximately;

/ divided by, per;

* multiplied by, and.

 

Some important ideas here are:

1) Temperature is a measure of a kind of energy – heat.

2) Increasing the temperature of a standard mass (e.g: 1 kilogram) of a substance a standard amount (e.g: 1 degree Celsius) requires a certain amount of energy. The ratio of energy to mass and temperature change is called the substance’s specific heat.

3) Ignoring complicating factors like heat loss due to imperfect insulation, it doesn’t matter what order one adds energy, mixes ingredients, etc.

3) If all ingredients are in the same phase – liquid, solid, gass – they all trade heat energy similarly, so ignoring weird exceptions, all ingredients in a mixture will eventually have the same temperature.

 

Hope this helps.