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.