Question

What is the heat of a reaction with a total reaction mixture volume of 70 5 mL if the reaction causes a temperature change of 6.4 °C in a calorimeter? Assume that the reaction mixture has a density of 1.00 p/ml. and a specific heat of 4.184 JigC The calorimeter has a heat capacity of 10.0 JC Answer

Answer 1

These can all be solved using the same expression:

$q=m\cdot c\cdot \Delta T$

Where 1 is the heat, m is the mass of the system, c is the heat capacity and delta T is the change in temperature (T_final - T_initial).

In the first problem (considering the print screens), we have 95.0 grams of a solution with a heat capacity of 4.184 J/g.K, with a change in temperature of 2.5 °C, which is equal to a change of 2.5 K (even though the final and initial temperatures are not the same in Kelvin, the difference is). So the heat is:

$q=95.0g\cdot 4.184\frac{J}{g\cdot ^{o}K}\cdot 2.5K=993.7J$

BUT this is the heat received by the solution. The heat "given" by the reaction is the same, but with the opposed sign; that is: -993.7 J.

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The second screen print is the same, but adding the heat capacity of the calorimeter, which must be added to the heat calculation:

$q=(m\cdot c+c_{calorimeter})\cdot \Delta T$

$q=(70.5g\cdot 4.184\frac{J}{g\cdot ^{o}C}+10.0\frac{J}{^{o}C})\cdot 6.4^{o}C=1952J$

(Where we have used that the density is 1.00 g/mL to calculate the mass, which is numerically identical to the volume).

Again, the heat of reaction is the same number, with the changed sign: -1952 J.

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Here we will use our heat equation for both the water and the metal, and take advantage of the fact that the heat "given" by the metal is equal to the heat "received" by the water:

$q_{metal}=-q_{water}$

$m_{metal}\cdot c_{metal}\cdot \Delta T_{metal}=-m_{water}\cdot c_{water}\cdot \Delta T_{water}$

The delta T for the metal is: 23.7 - 96.6 = -72.9 °C

And for water: 23.7 - 17.3 = 6.4 °C

We can rearrange the equation to calculate the c_metal:

$c_{metal} =-\frac{m_{water}\cdot c_{water}\cdot \Delta T_{water}}{m_{metal}\cdot\Delta T_{metal}}=-\frac{100.0g\cdot 4.184\frac{J}{g\cdot ^{o}C}\cdot 6.4^{o}C}{30.4g\cdot (-72.9^{o}C)}=1.21\frac{J}{g\cdot ^{o}C}$