Question

i need help for the rest

19.323 99.323 99.323 Plass of nickel Volume of water used (mL) Man of water (9) Iritual Temperature of Nickel Cami (Nicket Cams tend before transfere) 64.335 64,697 63.576 90 90 90 19 zo zo wale Caaled temp before transfer) (c) timal Temperature of Nickel Cans (C) Fimal temperature of Wales (c) 29 29 Tag 29 29 28 - AT of water. (oC) Head absorbed by hale (9 T Heat heat by mickel Cans Gh lot of micke / Loireda average specific heat of nicke Icon Standard Deviation RSA

Answer 1

lo calculate the Isulate the AT of Nickel con. I have to calculate the difference between final temperature of the after trans for and initial temperature before transfee Nicked coin in (6) AT = Tf - Ti - 29°C - 90°C 2-61c Nicked coin decrease 61°C AT of water is calculated as the difference between final temperature of the mass of water and the initial temperature of the same mass of water ATHO 29°C -20°C 9°C Water increase 9°C The AT of the coin in the triplicate replica 1 2 3 ATNI -61°C -616 -61°C ATWAO 10°C 9°C 18°C The Heat absorbed by water is calculated by next equation an Q = mazo o Ceno. AT Qu= 64,335gxCs, then(-646) - 4.18 Ja Piret Cesto = 10 cal = 4.18 TC

Centro Ce is soealic heat it is the amount of energy needed to increase the temperature of 4.09 of water from 14.5°C to 45.5°C. I suppose Censo does not depend on the temperature. Q = 64.3359.4.18 J . 10°C 2689.203 J Q2 = 64 90C 2433.901J .8°C = 2.125.981J PILIPILIIT To calculate heat take to the coin I have to equal it to the opposite of the heat absorbed by water Quse + Qui 20 for energy conservation Qui - Q H2O Q Ni first replica is Qm; 3 - Q, PPD- Q No - 2689.203J In second replica Qui = -Qz arrer QNi -2433.901J

Third replica QN, = -Q3 QNI -2425.984J Then, I am going to calculate specific hea of Nicked doindo Quoi = mwie Ce, Ni s ATNI 2 - Q H2O -Q1 = Man Ce, Nitx ATNI - 2689.203 J 99.3239. Ce, Mile (-610 -2689.203 J - Ceni 99.3239.(-64°C) 0.444 J. Ce wise -Q2 = 91.3639 r Cews, de a Torino -2433.901 J = 99.363gr Ce, wizef-61°c) -2433.901 J 20.402 - Ce, Magz 99.363946616) go - Q 3 = 99.3239 r Co, Nig3x (-64C) 22-2425.981 J_ Ce, M1,3 = 0.351 J. 99.3239 « (-616)

The average of Nick three replicas. The average of Nickel coin specific heat is only one; there is the average among three repli Cenis = 0.444 J Ce ne 2 = 0.40 275 Ce, Ni, 3 = 0.358 5 co Ceno, 1+ Ce, mi, z + Co, Ni, 3 Ce, Ni= 0.399 J 5 = 2,169.10 T S = 0.047 2 (Ce, ,n - CeNi) n is the amount of data averaged

Replica	1	2	3
$\dpi{80} \Delta T$ Nickel coin in oC	-61	-61	-61
$\dpi{80} \Delta T$ water in oC	10	9	8
Q1 (heat absorbed by water) (J)	2689.203	2433.901	2125.981
Heat absorbed by Nickel coin (J)	-2689.203	-2433.901	-2125.981
Specific heat of Nickel (J/(oC.g))	0.444	0.402	0.351
Average of specific heat of Nickel (J/(oC.g))	0.399
Standard deviation (J/(oC.g))	0.047

RSD (J/(oC.g))

0.118

RSD is calculated as SD/average

$RSD=\frac{S_{Ce,Ni}}{average _{Ce,Ni}}$

$RSD=\frac{0.047\frac{J}{g^{o}C}}{0.399\frac{J}{g^{o}C}}=0.11\bar{8}$