Question

In a steel plant, a 25g block of iron at 175 ˚C is dropped into 1 L of water at 20 ˚C in an insulated flask at atmospheric pressure. Calculate the final temperature of the system. The specific heat of iron is given by Ĥ(J/g)=17.3T(˚C).

What reference temperature was used as a basis for the entalpy formula?

Calculate the final temperature of the flask contents, assuming that the process is adiabatic, negligible evaporation of water occurs, negligile heat is transferred to the flask wall, and the specific enthalpy of liquid water at 1 atm and the given temperature is that of the saturated liquid at the same temperature. ( Enegy Balance reduces to Q= ΔH ).

Answer #1

Q = m c ∆T

Q = quantity of heat in joules (J)

m = mass of the substance acting as the environment in

grams (g)

c = specific heat capacity (4.19 for H2O) in J/(g
^{o}C)

∆T = change in temperature = Tfinal - Tinitial in ^{o}C

Heat lost by iron = heat gained by water

**Specific Heat Capacity of Iron = 0.45 J/(g
^{o}C). I am doing calculation as you mentioned
17.3**

25 x (175 -Tf ) x 17.3 = 1000 x 4.18 x (Tf-20)

75687.5 - 432Tf = 4180Tf - 83600

75687.5 + 83600 = 432Tf + 4180Tf

159287.5 = 4612 Tf

Tf = 34.53

**Hence Final temperature of the system is 34.53
^{o}C**

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