Question

=-72-88 kJ/mol. 10. A reaction known to release 1.78kJ of heat takes place in a calorimeter containing 0.100L of solution and the temperature rose by 3.65°C. The calorimeter was then rinsed out and emptied. To the empty calorimeter was placed a small piece of calcium carbonate and 0.100L of dilute HCI was poured over it in the same calorimeter). The temperature of the calorimeter then rose by 3.57°C. What is the heat, q, in kJ released by the reaction? Remember to include the correct sign in your answer. 1pt Heat capacity of Calorimeter +1000 L Solution = S ections 1.78KJ0

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Answer 1

We first need to calculate the heat capacity of the calorimeter, which we can do with the first experiment, using the expression:

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

Where q is the heat, m is the mass, c is the heat capacity and delta T is the change in temperature. We can rearrange and calculate for c (we will assume the density is 1 g/mL, which won't affect the final result, since we will then use it again and it's "effect" will be cancelled overall):

$c=\frac{q}{m\cdot \Delta T}=\frac{1780J}{100g\cdot 3.65^{o}C}=4.88\frac{J}{g\cdot ^{o}C}$

Knowing the heat capacity of the calorimeter, we can now calculate the heat it receives from the reaction of calcium carbonate with HCl (we will asume the density of the solution is the same as before and that the mass of calcium carbonate is so small it can be considered negligible, since its mass is not given):

$q=m\cdot c\cdot \Delta T=100g\cdot 4.88\frac{J}{g\cdot ^{o}C}\cdot 3.57^{o}C=1740J$

This is the heat "received" by the calorimeter which, as per the energy conservation law, is the same as the energy released by the reaction, which has the same numerical value, but the opposed sign. So, the heat released by the reaction is -1.74 kJ.