A piece of iron (specific heat capacity = 0.452 J. g ^-1. Degree C^1) with a mas

Question

A piece of iron (specific heat capacity = 0.452 J. g ^-1. Degree C^1) with a mass of 54.3218 g is heated in a water bath until it reaches a temperature of 98.2 degreeC. The iron is then transferred into a calorimeter (nested Styrofoam cups) that contains 68.5314 g of water initially at 22.1 degreeC. The final temperature of the entire system is 27.5 degreeC. The specific heat capacity of water is 4.184 J. g^-1 degree C ^-1. Based on this information, determine the value of the calorimeter constant. C_cal When 60.0 mL of a 0.400 M solution of HNO_3 is combined with 60.0 mL of a 0.400 M solution of NaOH in the calorimeter described in question #1. the final temperature of the solution is measured to be 26.6 degreeC. The initial temperature of the solutions is 24.0 *C. Assuming the specific heat capacity of the final solution is 3.90 J g^-1 degreeC^-1 and the density of the final solution is 1.04 g/mL. calculate q_rxn. HNO_3(aq) + NaOH(aq) rightarrow H_2O(1) + NaNO_3(aq)

Explanation / Answer

Solution :-

Q1). 1

Using the data of the iron lets calculate the heat lost by iron

q= m*c*delta T

= 54.3218 g* 0.452 J per g C *( 98.2 C -27.5 C)

= - 1736 J

Now lets calculate the amount of the heat absorbed by water using the data of water

q= m*c*delta T

= 68.5314 g* 4.184 J per g C* (27.5 C – 22.1 C)

= 1548.4 J

Now lets find the amount of heat abosorbed by calorimter

q cal = q iron - q water

          = 1736 J – 1548.4 J

         = 187.6 J

Now lets calculate the calorimeter constant

C cal = q/ delta T

         = 187.6 J / (27.5 C – 22.1 C)

         = 34.74 J/oC

So the calorimeter constat is 34.74 J / oC

Q2)

Volumes of the solutions are additive

So total volume of the solution = 60 ml +60 ml = 120 ml

Density = 1.04 g/ml

Mass of solution = volume * density

                              = 120 ml * 1.04 g/ml

                               = 124.8 g

Change in temperature = 26.6 C – 24.0 c = 2.6 C

Heat capacity of solution = 3.90 J/gC

Now lets calculate the amount of the heat given off by the reaction

Q= m*c*delta T

   = 124.8 g * 3.90 J per g C * 2.6 C

    = 1265.47 J

Now lets calculate the moles of the reactant

Moles = molarity * volume in liter

Moles of HNO3 = 0.400 mol per L * 0.060 L = 0.024 mol

Moles of NaOH = 0.400 mol per L * 0.060 L = 0.024 mol

Mole ratio of the reactants is   1:1 and moles are same

Thefore

Delta H rxn = (1265.47 J * 1 mol / 0.024 mol )*(1 kJ/1000 J) = 52.73 kJ/mol

Since the temperature is rising means heat is given off so the reaction is exothermic

Therefore we use negative sign to write the delta H rxn

So the delta H rxn = - 52.73 kJ/mol

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