Question

Free Response: Show your approach (i.e. proper equations). 25. 100 g of water( initially at 80°C) is combined with 80 g of cold water (initially at 0°C). Assuming that no heat escapes to the surrounding, and it occurred at a constant pressure, calculate: A) Final Temperature of the mixture B) AH of hot water and AH of cold water (after it reached Tfinal)

Answer 1

The solution is given below.

According to Principle of Calorimetry, Heat lost by a hot Heat gained by a cold 1 body body. Mars of hot water = 100g (mi) Initial Temperature (Tin) = 80°c of hot water Mars of cold water = 80g (m2) Initial Temperature (Tic) = 0°c of cold water (lic) = 0 Let the Temperature at equilibrium be x Then aty of hot water = (X-80) T: 072 7 Final Temperature And at of cold wates = (2-0) | Initial Temperature At ealibu ит, Heat lost by hot = Heat gained by cold wates -9 39 . -msat = m₂8ATz - mot, - MOTz Final Temperature of minture 1 - Temperature at equilibuicem - [100(4-80)]= 801x) | = 44.40°C -100% + 8000 = 80% . DH of hot wates = m sota 8000 = 80x+100 x - 100 X 4.184 x(80-44:44 8000 = 1807 100 X .184 X 35.56 na 800 - 148783 C 180 oh of cold water = m₂ SOTZ = 80 x4.184 x (0+44:44) = 20 *4.184 * 44.44 x = 44.44°C = 14874.95683 Watu da