lab2 write-up Feb26

On 2/26 class, we are given a question at the beginning of the class. Will the heat ball and ring shrink or expand? We drew the atom graph of the ball. Our group suspected the outside will expand outward, and inside will expand inward. Actually we find the hole doesn't become smaller as we assumed. A question is raised up: why the outward expand faster than inward? We assumed a conclusion, but we don't know if it is correct.

Then we solved a problem about thermal expansion: when we heat up the metal bar, what variables the final length will depend on? 1. initial length. 2.  material 3. temperature Tf-Ti.  The property of the material is represented as alpha. alpha=(deltaL/Li)/deltaT. This is called the coefficient of thermal expansion. likewise, we have another coefficient of thermal expansion of volume called Beta. which =(deltaV/Vi)/deltaT

Bimetallic strip is made of a strip of brass and invar. No matter we heat up invar side or brass side, the strip will bend to the invar side. on the contrary, if we put strip into ice, the strip always bends to the brass side.

Then, we solve a question about linear thermal expansion.

I this value valid or not?

We need to propagate the uncertainty of the measured variables.

thus, we have the result:

Therefore, I would report the alpha as

Then, we heat up a mixture of water and ice for 5 minutes until boiling. And continue to boil for another 5 minutes. We assume the graph should look like this:

The logger pro show the actual graph, which is not what we have predicted.

A question is raised; why did it stay constant? why does it have a small slope at the beginning?

The graph shows us: speeding up the molecules and breaking bond. 

Given the slope and power, we can calculate the mass of the mixture by deducting the formula Q=mc deltaT. We get m=177.7g

By given the conditions: delta m=80g. delta t=60s power=292w

We can calculate the latent heat of the system.

Graph shows the breaking bonds in ice is easier than in steam.

We conclude that heat flows from hot to cold.

We solve a problem: add 215kJ energy to 790g ice @ -5 degree, and find the final temperature.

The energy does not completely melt the ice.

We solve another problem: There is 255g ice at -12 degree. How much water at 22 degree we need to add to vessel to make the ice get to 0 degree?

Then, we are introduced to the ideal gas law. 

We did our first experiment, which is calculating the pressure by measuring the displacement of the water movement in the tube..

Here is the video taken while doing the experiment:

From P=F/A, we have concluded P=hpg in this case. (The deduction process is as shown below on the left side of the white board)

Then, we plug in the measured values and compute the pressure=539 pascals.

lab1 write-up Feb23

On 2/23 class, we first got familiar with unit of energy. 1 joule=1 N.m
1 joule=1000 kilowatts*seconds

Then, we solve a problem. The scenario is calculating the final temperature of a mixture of a cup of hot water and a cup of cold water. We use the formula Qc+Qh=0 to solve the final temperature.

Then , we put cold water inside the aluminum can, and put the can into the hot water. We have known the mass of 2 cups of water and the mass of the can. We also know both the initial temperature of hot water and cold water. We also know the final temperature of the system.
We use Q1+Q2+Q3=0 to solve the aluminum can's specific heat=16( unit here) Why would the experience off? We suspect several variables which the rate of cooling depend on, such as conducting rate alpha.

This is the graph from logger pro. We see how two systems get to equilibrium thermally. The final temp is around 30 degree Celsius. 

We suspect some reasons Ccan=16 is off. 

We know the rate of cooling depends on some variables . And the formula is Q=KA(Th-Tc)/L. 

This is an example in which we applied the formula and calculate

Then, we put a hot metal into the cold water. We find the relation between the released heat of the metal and the temperature change of the water caused by the metal. And by observing the graph, we find Q over T is a linear relation.