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Heat Effects and Calorimetry Essay

The first law of thermodynamics states that the energy of the universe is constant. Heat is transferred when the atoms of one material vibrate and collide with the atoms of another material, thus transferring energy. For this reason, heats always travel from hot to cold objects and two objects will reach an equilibrium temperature. Materials and Methods 1. Oven7. Test Tube 2. Pasco Temperature Probe8. Stopper 3. Scale9. Calorimeter 4. Samples of Metals10. HCl solution 5. Water11. NaOH solution 6. Beaker The sample metals are put to boil. In the first trial, the investigators use sample metal #2.

Pour an amount of distilled water in the beaker and measure. After obtaining the mass/weight of the distilled water, record its temperature using the temperature probe. The distilled water’s temperature (room temperature) is the initial temperature (Ti) of the water. Note that the initial temperature and the final temperature will be used to get the “change of temperature” of the water. Take out the metal from the oven and quickly put it in the beaker with water while simultaneously taking its temperature using the temperature probe. In this process, heat transfer is taking place.

While the computer is recording the measurement, changes in the temperature is evident. The water temperature goes up from its initial room temperature to a higher temperature until it reaches its equilibrium. Equilibrium is when the temperature is steady without any change. Weigh the metal and record the data then calculate. The heat of solution reaction is similar to that is present when a hot metal is put into water. There is an exchange of heat between the reaction mixture and the solvent, water. The heat flow associated with the reaction mixture is also equal to the enthalpy change.

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One of the simplest reactions that can be studied in solution occurs when a solid is dissolved in water. Same with the neutralization of HCl solution and NaOH solution, reaction occurs when solutions are mixed. Neutralization reaction occurs when an acidic solution is mixed with one that is basic. The heat effect will be measured when a solution of HCl, hydrochloride acid, is mixed with one containing NaOH, sodium hydroxide, which is basic. The heat effect is quite large, and is the result of the reaction between H+ ions in the HCl solution with OH- ions in the NaOH solution. Results

Obtain the necessary data before making the calculations. Heat lost=heat gained The specific heat of the water is 4. 184J/g °C which is equivalent to 1 cal/g °C. Sample Metal #2/Trial 1 The investigation starts with 100. 27 grams of water with the initial temperature of 23. 4°C. The temperature rises when the metal is dropped into the water to 25. 0°C. This means that the change of the temperature is 1. 6°C. The investigators know how many degrees the temperature of the water will rose. Investigators can calculate how much heat in joules will be transferred to the water from the metal.

Therefore, the calculation of the heat transferred to the water is: 4. 184 J/g °C x 100. 27 g x 1. 6 °C = 670. 6 J. The investigators calculate the specific heat of the metal after having the measurement of heat which was transferred to the water. The metal weighs 21. 28 grams. The temperature of the metal from the oven is 99. 7 °C and it cools from 99. 7 °C to 25. 0 °C thus, there is a change in temperature of 110. 4 °C. The calculation of specific heat of the metal is: Specific Heat of Metal = 670. 6 J____ 21. 28 g x 74. 7 °C =. 422 J/g °C Sample Metal #2/Trial 2

The temperature of the metal from the oven is 99. 7 °C. The metal was dropped into 47. 33 grams amount of water. The temperature reaches equilibrium as the final temperature at 28. 4 °C. The temperature change of the water is 3. 7 °C. Thus, the heat lost of the metal is equivalent to the heat gained by the water. The metal is weigh after cooling off and being recorded as: 4. 184 J/g °C x 47. 33 g x 3. 7 °C = 732. 7 J. The weight/mass of the metal is 29. 08 grams. The change of temperature is 92. 9 °C (99. 7 °C – 28. 4 °C = 71. 3 °C). Specific Heat of Metal = 732. 7 J___ 29. 08 g x 71. °C =. 353 J/g °C The solution reaction that we get is an endothermic because the change of solution came out to be negative with the solution of: Heat/g= __q___ = -991. 7= -199. 9 Mass salt 4. 96 The molar mass of the solution we use is 85. 01 (solid unknown number 22) and the heat solution per mole of compound we arrived to was -170. 01 KJ with the solution: 85. 01g x -199. 9 J x 1 KJ_ = -170. 01 KJ 1 g 1000 J The heat of neutralization, we arrive with change for the neutralization reaction of 5. 43 x10 J thus, the change of heat per mole of H+ and OH- ions reacting is 2. 7X10 KJ with the solution: 0. 025 x 40/ 1000= 2. 17×10 KJ. Conclusion In this experiment, the investigators use #2 metals. The #2 metal is probably copper since the specific heat of copper is 0. 385 J/g°C compares to our average specific heat of 0. 3875 J/g°C. The average molar mass of the metal we use is 65. 03 which is nearer to Zinc which has a molar mass of 65. 39 compare to our conclusion copper 63. 55. Considering the environment, heat lost is assumed during the transfer of metal from the oven to the investigators station. Beaker is used for safety reasons instead of Styrofoam cups.

Also, there maybe some discrepancy in the measurements of temperature of the metals due to the fluctuation of heat while putting in and taking out of the metals. This is probably one of the reasons why there is a lot of inconsistency in the results from the other investigators. Finally, in this kind of investigation, coming up with an accurate result will be possible only if the investigators has a clear guidelines to follow through. A lot of mistakes and assumptions are made on the process that leads to inaccuracy of results. Because of the limitations and mistakes therefore, recommendations to the mayor may seem inappropriate at this time.

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