Page 1 of 1 Chem 143 Specific Heat Capacity Lab Report. 4.) Calculate the change in temperature of the metal shot, A ATwer- ATtea= Tana-Tatia= (A6, B6, C6)–(A5, B5, C5). A6=272 Blo=24.4 co= 23.5 ATmeal (trial 1)= (A9) ATmea (trial 2)= (B9) A5=98.2 65=100 3 C5ニ100.7 ATsa (trial 3)= (C9) 5.) Now calculate the specific heat of your unknown metal using equation 4. Note the negative sign in Equation 4! Your specific heat capacity must be positive! 9water mmetal ATmetal A8 = 985.46J B8 = 103417j A2=B6.301 (Equation 4) Smetal = Here q.ater = (A8, B8, C8), mmstal = (A2), and ATmtal=(A9, B9, C9). Smuta(trial 1) = (A10) Smetal(trial 2) = (B10) Snta(trial 3) = (C10) 3 of 4

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Page 1 of 1
Chem 143
Specific Heat Capacity Lab Report
4.) Calculate the change in temperature of the metal shot, ATWater.
ATMetal= Tinal-Tinitial= (A6, B6, C6) (A5, B5, C5).
A6=27 2
Blo=24.4
co=23.5
ATMetal (trial 1)=
(A9)
ATMetal (trial 2)=
(B9)
A5=98.2
85-100 3
C5ニ100.7
ATMetal (trial 3)=
|(C9)
5.) Now calculate the specific heat of your unknown metal using equation 4. Note the
negative sign in Equation 4! Your specific heat capacity must be positive!
qwater
ATmetal
A8=985.40J
B8=1039.1
A2=36.301
Smetal
(Equation 4)
mmetal
Here
qwater
(A8, B8, C8), mmetal = (A2), and ATmetal=(A9, B9, C9).
%3D
Smetai(trial 1)% =
(A10)
Smetal (trial 2) =
(B10)
Smetal (trial 3) =
(C10)
3 of 4
14
Transcribed Image Text:Page 1 of 1 Chem 143 Specific Heat Capacity Lab Report 4.) Calculate the change in temperature of the metal shot, ATWater. ATMetal= Tinal-Tinitial= (A6, B6, C6) (A5, B5, C5). A6=27 2 Blo=24.4 co=23.5 ATMetal (trial 1)= (A9) ATMetal (trial 2)= (B9) A5=98.2 85-100 3 C5ニ100.7 ATMetal (trial 3)= |(C9) 5.) Now calculate the specific heat of your unknown metal using equation 4. Note the negative sign in Equation 4! Your specific heat capacity must be positive! qwater ATmetal A8=985.40J B8=1039.1 A2=36.301 Smetal (Equation 4) mmetal Here qwater (A8, B8, C8), mmetal = (A2), and ATmetal=(A9, B9, C9). %3D Smetai(trial 1)% = (A10) Smetal (trial 2) = (B10) Smetal (trial 3) = (C10) 3 of 4 14
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