Hand Warm Lab Report

The temperature probe was then quickly cooled to room temperature. When this was achieved, the hot water was immediately transferred into the calorimeter. This method of keeping the temperature probe cooled before measuring a new temperature was repeated throughout the entire experiment. Temperature data was collected for 180 s while swirling the temperature inside the calorimeter. The calorimeter still contained the warm water.

Use the following data to find the energy change in J of each system Specific heat of ice= 2.03 J/gC Specific heat of steam= 1.99 J/gC Specific heat of water= 4.18

Elijah Brycth B. Jarlos IX-Argon 1. Multicellularity is a condition of an organism to have multicellular cells. An example of a organism who has multicellular cells are plants, animals, and humans. The main reason of why scientists have a hard time finding a good set of existing organisms to compare. Is neither the first set of organisms which is being compared is dying as fast as the second specimen is being examined or they just can’t find the right species.

Shifa Sayeed can you check if this is all in past tense and if it has personal pronouns? The purpose of this experiment was to observe and thoroughly analyze how different substances of dissimilar intermolecular forces acted in different scenarios of evaporation, evaporative cooling, and boiling. In the lab, the three substances tested and compared were Acetone, Acetic Acid, and Propanol.

Experiment 1: Materials: • Alka-Seltzer tablets • Empty and clean water or soda bottles (12 oz to 24 oz) • Balloons • Water • Clock • Stove top Procedure: 1. Pour a sufficient amount of water (about 16 oz) into a small pot and place on the stove at high heat. 2. Watch the clock and after 30 seconds take the water off the heat.

Experimentally the energy gained/lose during the reaction was (INSERT ENTHALPY ANSWER). However, this number isn’t entirely correct because the heat loss in air wasn’t taken account for. A solution to this error could have been covering the Styrofoam cup.

Then the scientist will observe the different rates of reaction with temperature. The Boltzmann distribution of law, indicates that high temperature makes molecules gain high energy contents (pubs.acs.org/doi/abs/10.1021/ja). In order to measure the reaction rate, the scientists must use the same volume of water at three different starting temperatures: hot tap

Utilizing the method that the experimenters have come up with, the procedure isolates the variable being tested, which is the paper towel brand. By not changing other aspects of the procedure, other factors have a lower chance of affecting the accurate results of the answer to the main question. Additionally, the method is a quick and simple experiment, causing no use of excessive time than the provided duration. Within the first minutes of the experiment, the experimenters and the advisor realized that the method did not really provide accurate results due to merely estimating the numbers without the use of a graduated cylinder. Changing that aspect of the procedure by using a graduated cylinder to measure the amounts of water instead helped

They tested how the temperature would affect the rate of reaction. This was observed by the amount of time it took for the solution to change colors. For many chemical reactions there is an optimum temperature at which the chemicals will react with each other. As was found in their experiment, the temperature affected the rate of reaction. (Deoudes, 2010).

Introduction The intent of this experiment is to understand how hot and cold water interact with each other by combining clear hot water and black ice cold water. I hope to learn more about how hot and cold water interact with each other. As of now, I know that cold water is denser than hot water. Knowing this I formed my hypothesis.

In the warmth exchangers there are normally no outer responses to warmth and work. Down to earth applications incorporate worries about warming or cooling fluid vanishing or buildup of one or different liquid streams. In different applications, the objective may be reestablished or expel warmth or sanitization, purification and division, refining, fixation, crystallization, or control of the procedure liquid. In a few gadgets, and liquids that trade heat in direct contact. The vast majority of the warmth exchanger, and method for warmth exchange between the divider and Bond or transient way.

We know that energy is constant therefore any heat lost by our reaction is transferred to the surroundings. I was not able to locate any literature values for the change in enthalpy, despite looking very extensively. Error Analysis: For this experiment I assumed that the specific heat capacity for all solutions was 3.853 J/g°C.

Consequently, it was discovered that the aluminium can with cotton wool that acted as an insulator was discovered to be the best material as the final result was 69.5°C.7 This differed to the aluminium can without material as an insulator with a temperature of 62°C at nine minutes. The can with aluminium foil started off with a temperature of 73.5°C and slowly made it’s way to 65°C, proceeded it to be the second best insulator despite the material to have worked better at being a conductor. The can that had no materials was discovered to be the least efficient at trapping heat energy from hot water because one layer will not make a big difference as because aluminium is known to conduct heat better than insulating thermal

In calorimetry to find the amount of heat that was absorbed or released (q) by multiplying its mass (m), its specific heat capacity (c) and its change in thermal energy (∆T or Tf - Ti). The formula q= mc∆T is what was used in this experiment to determine the specific heat capacity of a small lead sinker. All substances are made up of particles that carry energy. The particles move faster when they contain thermal energy that is in the form of heat.

n=cV n=1.00 mol dm^(-3)×(25 dm^3±0.16%)/1000=0.025 mol±0.16 % The enthalpy of neutralization is then calculated. ∆H=(-1356.5 J±3.104% )/(0.025 mol±0.16%)=-54260 J 〖mol〗^(-1)±3.3 % ∆H=-54260 J m〖ol〗^(-1)±3.3%÷1000=-54 kJ 〖mol〗^(-1)±3.3 %