Thermochemistry Post-Lab Previous Question Score/Possible Points: No score possi

Question

Thermochemistry Post-Lab Previous Question Score/Possible Points: No score possible Total Score So FarTotal Exercise Points: 23/ 109 Scoring Scheme: 3-3-2-1 PartI For each trial, enter the amount of heat lost by the metal, metel Hint The specific heat of water is 4.184 Jig°C. Be careful of your algebraic sign here and remember that the change in temperature is equal to the linal temperature minus the intial temperature. Your answer should be reported to 4 digits Note: You should always carry 1 or 2 extra digits beyond the number of significant figures until your final calculation T metal 103.0 103.5 102.0 Masscool water T 69.638 70.291 70.299 Trial #MassMetal metal 183.399 183.399 183.399 23.0 23.0 23.0 36.5 36.0 36.5 #2: Quit Post-Lab

Explanation / Answer

For Trial 1:

Let the specific heat of metal be (qmetal)

Heat lost by metal = Heat gained by water

mass of metal * specific heat of metal * change in temperature =

mass of water * water specific heat* change in temperature =

183.399 * qmetal * (103-36.5) = 69.638 * 4.184 * (36.5-23)

q(metal) = 0.3225 J/gC

For Trial 2:

Let the specific heat of metal be (qmetal)

Heat lost by metal = Heat gained by water

mass of metal * specific heat of metal * change in temperature =

mass of water * water specific heat* change in temperature =

183.399 * qmetal * (103.5-36) = 70.291 * 4.184 * (36-23)

q(metal) = 0.3088 J/gC

For Trial 3:

Let the specific heat of metal be (qmetal)

Heat lost by metal = Heat gained by water

mass of metal * specific heat of metal * change in temperature =

mass of water * water specific heat* change in temperature =

183.399 * qmetal * (102-36.5) = 70.299 * 4.184 * (36.5-23)

q(metal) = 0.3305 J/gC

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