Question

Chemistry lab"Calorimetry"

Data: Insert data obtained from your TA here Mass of cups (g) 1 mass of cold water & caps (9) 15.1839 Temp. of cold water (-() Temp of I mass of inxed water mited CC water tapscg) 41.6 1 81.4882 5.6195 Calculations and questions: 21.3 Detailed approach for temperature determination Describe in detail the approach you have chosen to make the temperature determination. Clearly indicate why other available materials were not used in your method.

Answer 1

Assumption: since the weat capacity of the lup it not - mentioned in the quution, we will assume that the Cup' is insulator and no heat is consumed by the 'cup to raize its temperature We will use this approach that, the net heat given by hot water = the net wat gained by cold water. given Mals of cold water = 53.6839 -5.6695 = 48.olu u gouw Temp of cold water = 21.3°C . Mass of hot water = 81.4882-5 3.6839 = 27.8043 gram Final temp = uiroc het initial temperature of hot water be hoc : Heat gained by cold water heat given by hot - Meld x 8 x (Tq - To) d = m hut x B x (Tp-Tal mit -) 48 0144x(41.6-2113) = 27.8043 x (x - U16) - wate's

6 specific heat of water. (-48.0144 x( 20-3)= 27.8043 ( 1 – 41-6) » 85.05s4= n-Uk6 - [x = 76.6554oC Possible error: -lissince we neglected the weat gained by the cup , this will lead to error. (6) Also, these can be heat tranefes b/w water and the surrounding atmosphese which we neglected. : Estimated temp of warm water = 76.6554°C