Question

Connts Pd PartA Assume you ane perfomming the calbeacn step af Eperientand you begin wth 20 g of water at 20 Cand 20 gfwa 8C to be 45 Cht s the heatcpacty of the calorimtr Arddg the e pons ofwt yr ca p and long te procedure uined i the p t y d e e p d Assume room smperanane is 25C

Assume you are performing the calibration step of Experiment 8 and you begin with 20 g of water at 20 oC and 20 g of water at 80 oC. After adding the two portions of water into your calorimeter setup and following the procedure outlined in the experiment, you determine the temperature of the mixed portions of water to be 45 oC. What is the heat capacity of the calorimeter? Assume room temperature is 25 oC. Record answer in cal/degree celcius.

Answer 1

BMW solution Equation we are here are using q kalorimeter) I lib. ib. ibt e t -9 wam walen) = 1.600) werfen) q=mepot р - хресін? heet 07 - Change in temp m- muss Xa XA T. - m X Crowser AT = +mcpccoon $7+ Spectros) IN Ilmu 10 D -I coqm - 2000= ? calorimeter specific heat of water = 4.184 Elpam) . o Scalorimeter = – Prom - Seoul Go to 45-25) = - 4.184) 2 X doX(45-20) - 4.184 annex pocom) (u=Bch - 4184x 20 x - 4:184 x 2òn(-3-2 - - 20927. 2928.8 Cpx 20 =836.8 YY DDDD calon YUVVV CP celooimetaX 20 = 836.8 . I

(px 20 = 836.2 T 11 Cp = 41.84 Igre 5°C In calory ry 17 = 4.186 cal 2. cp = 418 a la cel foc (cp = 10 callo] the first change of temperature - (45-25) for is for calcrimefan. It is basically means that the calorimeten starts at room temp. and we have given that room temp =25°C.