A student performs the first trial of this experiment and...

A student performs the first trial of this experiment and obtains the following data. • 50.0 mL, 0.50 M CuSO4 • 5.0 mL, 5.0 M ethylenediamine (en) • Tinitial = 21.5°C. This is the average temperature of the CuSO4 and en solutions before mixing. • Tfinal = 27.5°C. The is the temperature value determine by plotting temperature vs time and extrapolating to t = 0 when the CuSO4 and en solutions are first mixed. • d = 1.0 g/mL • Csolution = Ssolution = 4.184 J/g °C • Ccalorimeter = 20.0 J/°C Calculate the heat of the reaction (Qreaction). Hint: See page 159 in the lab manual. Qreaction = -1.5 kJ/mol Qreaction = +1.5 kJ/mol Qreaction = -6.9 kJ/mol Qreaction = +6.9 kJ/mol

Transcript

00:01 So in this question, this is the metal.

00:04 So this particular metal, it is at high temperature.

00:08 So here what happens? it is transferring the metal, it is transferring his heat to the water.

00:17 So this particular water, it is at low temperature.

00:21 So here it is heat transferring from metal to the water.

00:25 So due to this, what happens? so it causes lower character.

00:31 Calculated value of specific heat of water so lower calculated value of specific heat of water how specific heat of water will see how so here what happens so the cue of the metal that is equal to that is a cue of the water plus heat loss you of the water plus heat loss so what is the cue here q it is nothing but heat transfer what is q it is nothing but heat transfer so here loss of heat so here heat it is loss getting loss that is here loss of heat so that what happens the q metal it gives lower value than actual it gives due to the heat transfer from the metal it gives lower value of lower value of lower value of specific heat lower value of specific heat than actual value so because here the cube metal is equal to because here q metal is equal to this mass of the metal so q metal is equal to mass of the metal q metal is equal to mass of the metal q metal is equal to mass of the metal into specific heat of the metal into delta t of the metal so what is delta t of the metal it is nothing but temperature change okay so here q metal decreases so q metal decreases because here heat transfer takes place so whenever we see that q metal it is directly proportional to the specific heat then what happens specific heat it gives the lower value it gives lower value that is the answer that is the answer then coming to the here next one so here in order to calculate the specific heat of metal specific heat of metal that is cp of the metal so here we have to go for the energy balance so we have to go for energy balance and also we have to consider no heat loss no heat loss so then only we can calculate the specific heat of the metal okay so m metal m metal specific heat of the metal and delta t of the metal that is equal to m water specific heat of water specific heat of water into delta t of water then only that means it indicates no heat loss or no heat gain so what happens here so the units here it is m of the m that will be ground moss will be in grams then specific eight jowls per gram degree centigrade delta t is equal to so here grams is equal to joules per gram degree centigrade that is delta c delta c delta c delta c delta c delta degree centigrade degree centigrade gram gram get cancelled out here also this a day so it becomes it becomes so first of all we have to we have to here q of water we have to find out...

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