Heat of Neutralization Lab Objective: The objective of this lab was to measure‚ using a calorimeter‚ the energy changes accompanying neutralization reactions. Background: Chemical changes are always accompanied by a change in energy‚ typically as heat. If the reaction releases heat (ΔH < 0) then the reaction is exothermic. If the reaction absorbs heat (ΔH > 0) then the reaction is endothermic. The quantity of heat is measured experimentally by having the reaction take place in an insulated container called a
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In this lab‚ we will measure the heat of combustion‚ or calorimetry‚ of a candle and compare the found quantity with known values for other hydrocarbons. The calorific value is the total thermal energy released when a substance goes through complete combustion with oxygen. In order to achieve the purpose of this lab‚ we must first determine the mass of the tea candle. Then‚ we will determine our room temperature‚ measure about 100 mL of chilled water‚ and then pour the water into the given empty
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Specific Heat Lab Objective: Find the specific heat of the unknown metal given using calorimetry. Background Theory: In every reaction‚ energy is transferred between a system and its environment. A system encompasses the substances that are involved in a reaction‚ and everything else in the universe other than the system is called the environment. The standard SI unit of energy is Joules (J). Temperature is the level of excitement of the atoms in a substance. In most cases‚ energy is
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Calculations/Analysis Through this lab‚ we are now able to calculate the molar heat of combustion for paraffin‚ since we have the difference of the mass in candle before/after and the periodic table of elements (for converting g to moles of paraffin). Molar heat of combustion = (kJ of heat)/(mole of fuel) However‚ we do not know how much heat was released nor the mole of fuel (paraffin). In order to find the amount of heat released‚ we use the formula: g=mcΔT. Here‚ g represents the heat‚ m represents the mass
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Heat engine lab Intro: when an engine runs‚ it pumps pistons that move up and down and provide energy to the engine to it to go. These pistons move because of pressure and heat. This work done on the system is not only mechanical but its also thermodynamic. When a piston undergoes one full cycle its displacement is zero because it comes back to its resting place. This means that its net thermodynamic work to be done should also be zero‚ as well as its total internal energy. In order to test this
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MECHANICAL ENGINEERING 449 SENIOR LAB Test of a Heat Pump Submitted Submitted by: Submitted to: Executive Summary: The purpose of this experiment was to determine the performance values of a Hylton Air and Water Heat Pump System. The system uses refrigerant 134a and water as the working fluids. The power input of the system was measured. The rate of heat output and the coefficient of performance are
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(The molar heat capacity of liquid water is 75.4 J/mol ºC; the molar heat capacity of ice is 40.2 J/mol ºC. The molar heat of vaporization of water is 40.7 kJ/mol. The molar heat of fusion of water is 6.02 kJ/mol.) (A) 71.8 kJ (B) 419 kJ (C) 64.2 kJ (D) 64.6 kJ (E) 11‚620 J 2. Which physical property of a liquid is NOT the result of strong intermolecular forces? (A) high vapor pressure (B) high boiling point (C) high heat of fusion (D) high melting point (E) high heat of vaporization
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Lab14: Heat Capacity Lab #14 – Heat Capacity OBJECTIVES In this lab you will: Use your code from the previous week to calculate the specific heat capacity per atom of a block of aluminum containing 35 atoms. Compare your theoretical curve with experimental data‚ and explain any discrepancies. Add code to your program to calculate the heat capacity of a block of lead containing 35 atoms‚ compare the theoretical curve with experimental data‚ and explain any discrepancies. Heat capacity is a measure
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Calculate the energy change (q) of the surroundings (water) using the enthalpy equation qwater = m × c × ΔT. We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. The water has absorbed the heat of the metal. So‚ qwater = qmetal Using the formula qmetal = m × c × ΔT‚ calculate the specific heat of the metal. Use the data from your experiment for the metal in your calculation. Aluminum: q=26*4.18*6.3=684.684‚ c=q/(m*^T)=3.913 Zinc:
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equation qwater = m × c × ΔT. We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. The water has absorbed the heat of the metal. So‚ qwater = qmetal m=mass of water= density x volume = 1 x 26 = 26 grams ΔT = T(mix) – T(water) = 38.9 – 25.3 = 13.6 q(water) =26 x 13.6 x 4.18 q(water) = 1478 Joules 2. Using the formula qmetal = m × c × ΔT‚ calculate the specific heat of the metal. Use the data from your experiment for the metal
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