Question

. A) 1.25 L B) 2.50 L Q13. 1.0 L of 1 M NaOH(aq) and 1.0 L of 1 M HNO3(aq), both initially at 25.0 °C, were mixed calorimeter. The temperature of the solution rose to 28.1°C. How much heat was released? Assume the specific heat capacity of all solutions is 4.184J/(g°C), and the density of all solution is 1.00 g/l A)-80.1 kJ B) -77.8 C)-52.0 l D) -26,0 11

Answer 1

The formula to calculate the heat is:

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

Where q is the heat, m is the mass, c is the heat capacity and delta T is the change in temperature.

In our system, we have a total of 2 L, which are equivalent to 2 kg, due to the density being 1.00 g/mL. c is also known and delta T is: 28.1 °C - 25.0 °C = 3.1 °C. So, the heat is:

$q=2000g\cdot 4.184\frac{J}{g^{o}C}\cdot 3.1^{o}C=25941J\cong 26.0kJ$

This is the heat absorbed by the solution, which is the same as the heat released by the reaction, which has the same absolute value, but with a negative sign: -26.0 kJ.