2/26/2015 week1 Day2 Thermal Expansion and Latent Heats

this photo is that at the same time heatting a ring and a ball, the atom's graph. The atoms of ring come to outside and inside at the same time, it should be increasing of inside and outside diameters, but the atoms of the outside diameter has the more power to pull the inside diameter, so in fact, the inside and outside diameters are both increased. For the bll, heating it that makes the atoms go outside that makes the ball bigger.

When only heating the ring, like we talk about in the first photo, the ring's inside and outside diameters are increasing at the same time. So the ball can cross the ring.

then we talk about the 4 reasons that can make the bar bigger. and find the unit of α is (℃)^(-1).

these three photos are the experiment that if we heat a thing that one side is brass and the other side is invar, what will happen. and the first photo is that before the experiment that we guess that it will curve to the right. because the left is brass and brass's coefficient of volume expansion is larger. then the professor heats right side and the left side one by one and the ending is that it always curve to the right. 

then the professor put it in to the ice water , we guess this that it will curve to the left and the result is same like our guess. so we can know that because the coefficient of volume expansion depends that which side will curve.

 

then we begin an experiment about the linear thermal expansion. the first four photos are the equipment. we use vapor to heat the tube and use logger pro to find the coefficients of linear expansion of the tube.

We used the data of the ΔL, Lo, r, ΔT and θ to find the α of the aluminum rod.this is the process of calculating,we use the equation on the right side and the sitar we find in the logger pro to calculate the coefficient of linear expansion of the tube is 1.9*10^(-6)(℃)^(-1）.

this experiment is that put the ice in the water and use the equipment to heat for 5mins. the second photo is the graph about this experiment. the graph of the third photo is that our guess about the graph.

and the equations are that we find the m of the water if we use a power of 293W to heat water by 300s from 0 ℃ to 100℃， the slope is dT/dt IS 0.3495,and the matter is 209g.

these two photos are to find the pressure of the water if we are blowing the water to raise a certain height. we use the equation at the top of the photo.

this photo is to figue out that if there are ice and water in the same place, how much energy that need to use to heat to a special temperature. at that time, we can not forget that the energy of ice becoming water, and we use the equation: Q=mL. L is the heats of fusion or vaporization.

this is another example that to find how much ice we need to make 255g water from 12 ℃ to 0℃.

Conclusion:

Today in class, we focused on  the thermal expansion and latent heat and the idea of thermal expansion. The change in temperature cause objects to expand by doing  letting the ball go through, heating the aluminum rod and calculating its expansion in length, heating water mixing with ice.We did activities such as heating the steel ring to and to calculate the mass of water in the cup, doing some latent heat problems. The  melting or evaporation change happen in ice and water that is significant in determining the heat gained equal to the heat lost idea in changing systems. we learn the ideal gas law pv=nRT by using pressure, volume, and temperature. 

2/24/2015 week 1 day1 of class Teamperature and Heat

This lab began with the definitions of the basic units of measurement. Conversions between Fahrenheit and Celsius were made. Below shown is the visual representation of such.

every group uses the equation: K=273+Celsius to find the temperature of room. and the ave of temperature is 295.33. then we use the equation at the bottom of photo to find the uncertainty. and the uncertainty is 1.72.

professor gives us a cup of hot water and a cup of cold water. because we know the heat does not change, and only transfer from hot one to cold one, so we use equation Q=cm(T_f-T_i) to find the final temperature of the mix water.

this photo is that we do not use any number. just the letter to express the cold water and hot water, afer solving the equation, we use numbers to find the final temperature.

this time we use a can to install the water.the equation is same one : Q_increase+Q_decrease=0, and Q=c*m*(T_f-T_i)

the first euqation is about how to transfer from kwh to J. and the second part is about what is the distant that use 9.81N to do 1J work. 

this photo is about the hear transfer. . dQ/dt IS the heat current. k is the thermal conductivity of the material. the quantity (T_H-T_C)/L is the temperature difference per unit length; it is called the magnitude of the temperature gradient. T_H and T_C are the temperatures on the two sides of the slab. R is L/k is 1m^2*K/W.

this photo is about the reasons to influence the heat escape.

this photo is about thre rate of cooling.

this is an example of H. and we should find the final temperature of Al. 

these two photos are about an example about use the linear to find the specific heat.

Conclusion:
Today in class, we analyze temperature and heat and the process of thermal conductivity through heat energy, flow,and conduction. We learn conversions of temperature in Kelvin, Celsius, and Fahrenheit and their relationship. In fact thermal resistance R was found to act as a thermal resistance where as the R increased, the heat flow decreased. The heat conductivity modeled with two joined bars of copper and aluminum yielded different values that reveals how there is more heat loss in aluminum compared to copper. The type of material and its heat capacity is a factor that determines its heat flow. In experiment we use a mixing process with an aluminum can  that showed two curves at opposite ends that reached equilibrium in temperature.and we know heat conductivity has the same flow rate. And we learn the specific heat in heat energy and the relationship of heat in temperature and joules. we learn formulas  such as dQ/dt = A (Tf - Ti) / R, dQ/dt = kA (Tf - Ti) / L, and R = L/K.