Question

please help! its due tomorrow!!!

For any problem that requires a calculation, work must be shown to POEL 1. A student collected the following data set for this lab. Calculate the following quantities Temperature of 50.0 mL of warmer water: 43.0°C Temperature of 50.0 mL of cooler water: 22.9°C Temperature after mixing: 32.7°C a. Heat lost by the warmer water b. Heat gained by the cooler water c. Heat lost to the calorimeter d. Heat capacity of the calorimeter

Answer 1

Warmy 3=1. 6 warmer water is 50.0 ml=0 mass = 50.0x! = 50 g, Temperatures 43.0°c Cooler wafer = 50 g, Tember afure - 22.9°c final temperature = 32.7°C. @ Heat Lost of warmer wafer = msit mato g . - 50X1X (43- 32.7) TT 2 9, = 515 cal, © Heat gained by Cooler coafes - amsat = 50x1(32.4-22:3) cal 8 L = 490 cal, B now 1. 1 Heat gaine Heat host I cool water efficat que calometfre a warm water S o , ot's to get - How Heat gain = Heat Lost cool writer & calomimefor a warm water... 92 + g = 8 8 = 9,-82 = 1515-490) Cal 2 25 cal, • 20 Heat Lost lue to calorimeter is a5 cal, @ Heat capaeity of Calorimeter is = ep. I now Heart dost = Cp dT =25 daz (32.7- 22:90 27 28 298 - 8.55; callkel co Herit capacity of Calometer is 2.55" alik, 25 -25 l: 2 998