(a) Find the heat capacity of your calorimeter if the final temperature was in f

Question

(a) Find the heat capacity of your calorimeter if the final temperature was in fact 0.4 degrees higher.

(b) Suppose the heat capacity of the calorimeter is in fact 10% less than the value determined in Part 1. What would be the impact of this error on the final result, i.e. on the enthalpy of formation of MgO (s)? To answer this question, simply recalculate enthalpies of two reactions, i.e. MgO + HCl and Mg + HCl, using the new value of the heat capacity of the calorimeter.

These are the exact way the questions are asked. I found this in a previous question. Maybe it will help. Ex. 1 Determine the heat capacity of calorimeter from the following data: mass of hot water 72.79 g, temp of hot water (and calorimeter) 59.2 degree C, mass of cold water 76.74 g, temp of cold water 21.2 degree C, and final temp. 41.7 Degree C. The assuption in this determination is that the heat absorbed by the cold water comes soley from the heat givien up by hot water and colorimeter.

Explanation / Answer

Mass of hot water, Mh = 72.79 g

Temperature of hot water, Th = 59.2 C

Let heat capacity of calorimeter be Cpc J/ C

Specific heat capacity of water, Cpw = 4.186 J/g-C

Mass of cold water, Mc = 76.74 g

Temperature of cold water, Tc = 21.2 C

Final temperature, T = 41.7 C

Heat balance:

Heat lost by hot water + calorimeter = Heat gained by cold water

Mh * Cpw * (Th – T) + Cpc * (Th – T) = Mc * Cpw * (T – Tc)

72.79 * 4.186 * (59.2 – 41.7) + Cpc * (59.2 – 41.7) = 76.74 * 4.186 * (41.7 – 21.2)

Cpc = 71.6 J/C

a)

If final temperature, T is 0.4 C higher

T = 41.7 + 0.4 = 42.1 C

Heat balance equation becomes

72.79 * 4.186 * (59.2 – 42.1) + Cpc * (59.2 – 42.1) = 76.74 * 4.186 * (42.1 – 21.2)

Cpc = 87.9 J/C

b)

Enthalpy calculated, Q = Energy change of water + Energy change of calorimeter

Q = M * Cpw * T + Cpc * T

Since Cpc is actually 10% less the than value calculated in part 1

Qactual should be less than Qcalculated

Actual enthalpy of formation will be less than calculated enthalpy.

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