Suppose you add 92 ml of dilute acid to 46 ml of water. Both the water and acid were at room temperature (22 C) before being mixed. The temperature of the solution increased to 39 C. How much energy (in kJ) was released by the dissolution of the acid if the specific heat capacity of the solution is still the same as water 4.18 J/(g ·K) and the density was still approximately 1.0 g/ml. The error interval is +/- 5%.
energy=mass*specificheat*changetemp
= (92+46)grams*4.18J/gC*17C
To find the amount of energy released by the dissolution of the acid, we can use the equation:
Q = m × ΔT × C
Where:
- Q is the amount of energy released or absorbed (in joules or J)
- m is the mass of the solution (in grams or g)
- ΔT is the change in temperature (in Celsius or °C)
- C is the specific heat capacity of the solution (in J/(g ·K))
We are given that the volume of the acid added is 92 ml and the volume of water is 46 ml. Since the density of the solution is 1.0 g/ml, the mass of the solution is equal to its volume. Therefore, the mass of the solution is 92 g + 46 g = 138 g.
The change in temperature is given as 39 °C - 22 °C = 17 °C.
Now we need to convert the mass of the solution from grams to kilograms (kg) and the temperature change from Celsius to Kelvin (K) because the specific heat capacity is given in J/(g·K).
Mass of the solution (in kg) = 138 g / 1000 = 0.138 kg
Temperature change (in K) = 17 °C + 273.15 = 290.15 K
Now, we can substitute the given values into the equation to find the amount of energy released:
Q = 0.138 kg × 290.15 K × 4.18 J/(g·K)
Calculating the value of Q gives us:
Q = 16.2255 J
However, we need to convert this value to kilojoules (kJ):
Q = 16.2255 J / 1000 = 0.0162255 kJ
Therefore, the amount of energy released by the dissolution of the acid is approximately 0.016 kJ.
Considering the error interval of +/- 5%, the final answer will be:
0.016 kJ ± (0.016 kJ × 0.05) = 0.016 kJ ± 0.0008 kJ