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Experiment to find the specific heat capacity of an aluminium block. DATA COLLECTION: Mass(m) (kg) (5dp) Current(I) (A) (2dp) Voltage (V) (2dp) Time(t) (s) (0dp) Initial temp(T1) (ºC) (2sf) Final temp(T2) (ºC) (2sf) +/-( 5*10^ -4) % +/- (1*10^ -1) % +/- (4*10^ -2) % +/- (5*10^-1)s +/- 0.5 ºC +/- (5*10^ -1) ºC 0.99577 3.70 12.07 0 16.0 16.0 0.99577 3.70 12.08 60 16.0 18.0 0.99577 3.68 12.10 120 16.0 20.0 0.99577 3.66 12.11 180 16.0 22.0 0.99577 3.68 12.10 240 16.0 25.0 0.99577 3.64 12.08 300 16.0 27.0 0.99577 3.64 12.08 360 16.0 30.0 0.99577 3.65 12.03 420 16.0 33.0 DATA PROCESSING AND PRESENTATION: As Q=m.c.(T2-T1) Where Q= energy transfer C= specific heat capacity We can rearrange this to give: C= Q/(m(T2-T1)) And as power = energy/time Therefore E= Pt = Q And P = IV therefore Q = IVt Hence C= IVt/(m(T2-T1)) Which is rearranged to the form y=px + c to give: T2= (IVt / (m.C)) + T1. Where p is the gradient, and equals 1/C, therefore x = IVt/m = Q/m, and y = T2 the y intercept is equal to T1 Therefore I have calculated this table: Energy transfer Errors (J) Q/m (j/kg) Errors(J/kg) Final temp(T2) ( ºC) (2sf) (Q) (J) (0dp) (0dp) (0dp) (0dp) +/- (5*10^ -1) ºC 0 +/- 0 0 +/- 0 16.0 2682 +/- 27 2693 +/- 30 18.0 5343 +/- 32 5366 +/- 30 20.0 7978 +/- 36 8012 +/- 40 22.0 10687 +/- 41 10732 +/- 40 25.0 13191 +/- 46 13247 +/- 50 27.0 15830 +/- 50 16897 +/- 50 30.0 18442 +/- 55 18520 +/- 60 33.0 The errors were calculated...

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