Data Ist Trisl (2nd Trfal Kind of netal Mass of metal Mass of cup Mass of cup an
ID: 558968 • Letter: D
Question
Data Ist Trisl (2nd Trfal Kind of netal Mass of metal Mass of cup Mass of cup and water Hass of water Temperature of hot netal Original temperature of water fn cup 1059 1as 9. Final temperature after addition of metal to water Calculations: SHOW ALL WORK FOR THE 1st TRIAL. Use unfts and significant figures 1. How much heat (oules)ves gafoed by the water? 2: How much heat was lost by the netal? ,· Determine the spéeTffè. helt of the tet 4. Give the set-up and value for the percernt error of your spegitic heat. 5. Give the set-up and value for the approxinate stonic wetght AW) of the netal using the Rule of Dufong and Petit and your experinental specific heat, Use unfts.Explanation / Answer
1. mass of water = 98.5 g
specific heat capacity of water = 4.184 J/g.oC
temperature change dT = 26.11 - 15.55 = 10.56 C
heat energy gained by water (q) = m x s x dT
=> q = 98.5 x 4.184 x 10.56
=> q = 4352 J
2. From calorimetry,
heat energy gained by water = heat energy lost by the metal
=> heat energy lost by the metal = 4352 J
3.
q = m x s x dT
mass of the metal = 125 g
specific heat capacity of the metal = s = ?
temperature change dT = 99 - 26.11 = 72.89 C
=> 4352 = 125 x s x 72.89
=> s = 0.478 J/g.C
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