Question

Please help me figure out the blank part from the 1st photo..temp of metal was 60 degreesc

61 Run l 8 30.000 g Run 2 Unknown humber Mass of mera 3 volume of woer (mt) 30.00 얹 25.00 m L 23.i'C 25.00mL Ini 141 dcrti (℃) Density of water 003/1009/Cm Temperatue o meral 4.184 Joul Jaules gonC Atw 67.5'C Euaigy gained by waser Specitic heot of un rown neal Avg .№lue of specfic hla+ if two nas Perun Atomic mss of unknown metal Pozable idanity of uniknown meta ) SIGNATURE DATE WITNESS/TA DATE THE HAYDEN-McNEIL STUDENT LAB NOTEBOOK NOTE:INSERT DVIDER UNDER COPY SHEET BR

Answer 1

In this experiment, the final temperature of the system is the maximum temperature reached by the system.
From the graphs,
final temperature of the system in run 1 = 33.1^oC
final temperature of the system in run 2 = 32.8^oC

Thus, temp. change in water = 33.1-21.1 = 12^oC (Run 1) and 32.8-23.1 = 9.7^oC (Run 2)

Energy gained by water = mass of water x Sp. heat of water x temp. change of water
Thus,

Energy gained by water (Run 1) = 25.0 x 4.184 x 11.9 = 1255.2 Joules gm^-1oC^-1
Energy gained by water (Run 2) = 25.0 x 4.184 x 9.7 = 1014.62 Joules g^-1oC-1

Now, Energy gained by water = energy released by the metal

$specific \, heat\, of\, metal=\frac{Energy \, released \, by\, the\, metal}{mass\, of \, metal\,\times temp.\, change\, of\, metal}$

Thus,
$specific \, heat\, of\, metal (Run \, 1)=\frac{1255.2}{30\times 67.5}=0.62 \frac{J}{gm\, _{\, }^{o}\textrm{C}}$
$specific \, heat\, of\, metal (Run \, 1)=\frac{1014.62}{30\times 67.7}=0.50 \frac{J}{gm\, _{\, }^{o}\textrm{C}}$

Average Sp. heat of metal = (0.50 + 0.62)/2 = 0.56 J gm^-1oC^-1

The given Sp. heat value correlates to the Sp. heat value of Iron metal. (⁵⁶Fe)

	Run 1	Run 2
Final temp. of the system (oC)	33.1	32.8
Change in temperature of water (oC)	12	9.7
Energy gained by water