A piece of iron (specific heat capacity = 0.452 J/g*C) with a mass of 54.3218g i

Question

A piece of iron (specific heat capacity = 0.452 J/g*C) with a mass of 54.3218g is heated in a water bath until it reaches a temperature of 98.2 C. The iron is then transferred into a calorimeter (nested styrofoam cups) that contains 68.5314g of water initially at 22.1 C. The final temperature of the enitre system is 27.5 C. The specfic heat capacity of water is 4.184 J/g*C. Bases on this information, determine the value of the calorimeter constant, Ccal.

When 60.0 mL of a0.400 M solution of HNO3 in 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 C. The initial temperature of the solutions is 24.0 C. Assuming the specific heat capacity of the final solution is 3.90 J/g*C and the density of the final solution is 1.04 g/mL, calculate qrxn + qsoln + qcal = 0.

HNO3(aq) + NaOH(aq) --> H2O(l) + NaNO3(aq)

*C meaning celsius

Explanation / Answer

Ans 1

Heat lost by iron = heat gained by calorimeter + heat gained by water

Q iron = Q cal + Q water

Mass of iron x Cp iron x change in temperature of iron = Ccal x change in temperature + mass of water x Cp water x change in temperature of water

54.3218 g x 0.452 J/gC x (98.2 - 27.5)C = Ccal x (27.5-22.1)C + 68.5314g x 4.184 J/gC x (27.5-22.1)C

1735.929 J = Ccal x 5.4 C + 1548.371 J

Ccal = 34.73 J/C

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