Question

< Question 10 of 19 > A hot lump of 30,4 g of copper at an initial temperature of 64.9 °C is placed in 50.0 mL, H,O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water, given that the specific heat of copper is 0.385 J/(g. °C)? Assume no heat is lost to surroundings. final = M

Answer 1

Dear student,

T_{​​​​final} = 27.11°C

His o The heat eneagy that is absorbed by the water is equal to the heat encagy that is absorbed by the incon o q = m C DT , where DT = Tf-Ti & waler & Vou =0 quater = -Vou Now V watcn = 50g *4. 084 5 1gic (Tf- 25°c) Vuater = 209-2 (1f- 25 c) Vwateen - 209:2 Tf - 5.2300 V = 3ong x 0.385 J/g °c (Tf - 64.9%) = 11.704 (Tf - 64.90) = 11.704 Tf - 759.58'C binco a water = -Vou 204277 - 5230L =-11.704 T7 + 739.58'( 2092 Tf +1.704 14 = 759.58*C + 5230°C 2 20.9 Tf = 5989.58°C Tt = 5989.58 220.9 = 27.11°C 2010 11 16

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