Experiment #1waterpipeinitial temperature25ðC100ðCfinal temperature28.7ðC 28.7ðCÃÂT3.7ðC-71.3ðCmass125 g60gÃÂT = Tfinal àTinitialhave two unknowns for the pipe â specific heat (c) and q-use water to calculate itqw = mwcwÃÂTwqw = (125 g) (4.14 J/g*ðC)(3.7ðC)qw = 1900 Jq is positive when energy is absorbedamount of energy absorbed is equal to the amount of energy releasedqwater = -qpipeqpipe = -1900 Jcpipe = qpipempipeÃÂTpipecpipe = -1900 J = 0.4 J/ g*ðC(60 g) (-71.3ðC)metal of the pipe â Fe (iron)Experiment #2ethanolpipeinitial temperature25ðC100ðCfinal temperature31ðC31ðCÃÂT6ðC-69ðCmass125 g60 gwant to find out specific heat of ÃÂethanolàâ not sure if it is ethanol-or whatever Dr. Rosenthal said (so its like solve it without looking at table)Step #1; use cpipe from experiment #1Step #2 calculate qpipe = mpipecpipe ÃÂTpipeStep #3 ; -qethanol = qpipeStep #4 cethanol = qethanolmethanol -->TethanolNaOH (s)--> NaOH (aq)bottom of the beaker felt hot â energy is being released by the system into the surroundingssystem: NaOH (s) Na+(aq) OH-(aq)surroundings: watertemperature of the water is increasing during the dissolving â energy is being released by the system and absorbed by the surroundings (water)-means that the kinetic energy of the water increased (kinetic energy is a reflection of temperature)-the chemical potential energy of the system decreased-released potential energyâ the ions are spreading apart (the relative positions of the ions changed)-any change in potential energy â called enthalpy change (in this case, exothermic àenergy is released)NaOH (s) â NaOH (aq)Potential Energy change = enthalpyEnthalpy àthe potential energy change that a system undergoes, reflected by a change in temperature of the surroundingsÃÂheat contentà(enthalpy) of substancesÃÂH â symbol for enthalpy ÃÂH = mcÃÂTheat content of the water (surroundings) increasesÃÂH is positive for waterÃÂH is negative for NaOHConclusions: exothermic reaction (negative ÃÂH for system â potential energy went down, released by the system)-energy is a product NaOH (s)...
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... heat of combustion also increases. This is because more reactants (in terms of atoms) form more products, yielding greater enthalpy. The result is consistent in both moles and grams. 4. Oxidation-reduction reactions are increasingly important as a source of energy. Explain the displacement of ...
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