A 19.6 g sample of a metal was heated to 67.1 degree C. When the metal was place

Question

A 19.6 g sample of a metal was heated to 67.1 degree C. When the metal was placed into 27.6 g of water in a calorimeter, the temperature of the water increased from 20.0 degree C to 35.0 degree C. What is the specific heat capacity of the metal? The specific heat of water is 4 184 J/(g degree C) From the enthalpies of reactions N_2 (g) + O_2 (g) rightarrow 2 NO (g) Delta H = 180.7 kJ 2 NO (g) + O_2 (g) rightarrow 2 NO_2 (g) Delta H = -113.1 kJ N_2 O (g) rightarrow N_2(g) + 1/2 O_2 (g) Delta H = 81.6 kJ Use Hess's law to calculate Delta H for the reaction N_2 O (g) + NO_2 (g) rightarrow 3 NO (g)

Explanation / Answer

m = 19.6 g

T = 67.1

m = 27.6 g w

Tw = 35-20

Cp of metal

Apply heat balances

Qm = Mm*Cpm*(Tf-Ti)m

Qm = 19.6*Cpm*(35-67.1)

Qw = Mw*CpW*(Tf-Ti)w

Qw = 19.6*4.18*(35-20)

Qm = -Qw

19.6*Cpm*(35-67.1) = -19.6*4.18*(35-20)

Cpm = (-19.6*4.18*(35-20))/(19.6)/(35-67.1) = 1.953271

Cp metal = 1.95

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