Gradebook -Print Calculator Periodic Table Question 14 of 27 Incorrect Map Sapli

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Gradebook -Print Calculator Periodic Table Question 14 of 27 Incorrect Map Sapling Learning A hot lump of 31.5 g of copper at an initial temperature of 53.7 is placed in 50.0 mL of H2O 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·)? Assume no heat is lost to surroundings Number 25.2 There is a hint available! View the hint by clicking on the bottom divider bar. Click on the divider bar again to hide the hint. Close Previous Give Up & View Solution O Check Answer Next Exit Hint

Explanation / Answer

1)   heat lost by copper = heat gained by water

mass of Cu*S1*DT1 = mass of water * S2*DT2

mass of Cu = 31.5 g

mass of water = v*d = 50*1 = 50 g

S1 = specific heat of copper = 0.385 j/g.c

S2 = specific heat of water = 4.18 j/g.c

(31.5*0.385*(53.7-x)) = (50*4.18*(x-25))

x = final temperature of mixture = 26.57 c

2)

heat released(q) = C*DT

     C = heat capacity of calorimeter = 30.3 kj/k

     DT = 2.13 c

q = 30.3*2.13

    = 64.54 kj

    = 15.42*10^3 cal

nutritional calories = (15.42*10^3)/3.3 = 4672.7 cal/g

3) No of mol of KSCN = w/Mwt = 5/97.2 = 0.0514 mol

heat absorbed(q) = C*DT

                   = 3.193*0.39

                   = 1.245 kj

DHsol = q/n = 1.245/0.0514 = 24.22 kj/mol

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