Question 3 a) (5 points) We place 50.0 g of a metal, that has a temperature of 9

Question

Question 3 a) (5 points) We place 50.0 g of a metal, that has a temperature of 90.00C, in 200.0g of water that has a temperature of 20.00°. The water is in a beaker that also has a temperature of 20.00 C. The beaker has a calorific capacity of 555 JK-1 and the specific heat capacity of water is 4.184J K1g1. The final temperature is 21.66 C. What is the metal's specific heat capacity. b) (3 points) The enthalpy of formation of liquid water is -285.8 kJ mol1 Water's enthalpy of condensation, H2O (g), is-44.0 kl mol-1, what is the value of ?? for the following reaction? The entire process occurs at 25.0°C 2H2 (g) + O2 (g) -> 2H20 (g)

Explanation / Answer

3)

a) heat lost by metal rod = heat gained by water + beaker

    m1*S1*DT1 = m2*S2*DT2 + c*DT

m1 = mass of metal rod = 50 g

s1 = specific heat of metal rod = ? j/k.g

DT1 = 90-21.66

m2 = mass of water = 200 g

s2 = specific heat of water = 4.184 j/k.g

DT2 = 21.66-20

C = calorific capacity of beaker = 555 j/k

DT3 = 21.66-20

50*x*(90-21.66) = 200*4.184*(21.66-20)+555*(21.66-20)

x = specific heat of metal= 0.676 j/g.c

b)

   from the given data,

    enthalpy of formation of H2O(g) = -285.8-(-44.0) = -241.8 kj

DHrxn = (2*DH0f,H2O(g))-( DH0f,H2(g)+DH0f,O2)

        = (2*-241.8)-(0+0)

        = -483.6 Kj

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