The density of a 1.0 M solution of NaCl is 1.04 g/mol and its specific heat, c, is 3.60 J/C*g. How much heat is absorbed by 31.4 mL of this solution if the temperature rises from 18.6 C to 29.7 C?
Use the density to convert 31.4 mL of the solution to mass. Then,
q = mass x specific heat x (Tfinal-Tinitial).
is it 0.6527 J?
I got 1310J
To determine the amount of heat absorbed by the solution, we can use the formula:
Q = m * c * ΔT
Where:
Q is the amount of heat absorbed or released (in Joules)
m is the mass of the solution (in grams)
c is the specific heat capacity of the solution (in Joules per gram per degree Celsius)
ΔT is the change in temperature (in degree Celsius)
First, let's calculate the mass of the solution. The density of a 1.0 M NaCl solution is given as 1.04 g/mol. Since 1 mole of NaCl is dissolved in 1 liter (1000 mL) of water to prepare a 1.0 M solution, we can find the mass of the solution using the molarity and density.
Mass of solution = molarity * molar mass * volume of solution
The molar mass of NaCl is 58.44 g/mol.
Mass of solution = 1.0 mol/L * 58.44 g/mol * 31.4 mL * (1 L / 1000 mL)
Mass of solution = 1.834 g
Now we can substitute the values into the heat equation:
Q = m * c * ΔT
Q = 1.834 g * 3.60 J/C*g * (29.7 C - 18.6 C)
Q = 1.834 g * 3.60 J/C*g * 11.1 C
Q = 71.56 J
Therefore, 31.4 mL of the 1.0 M NaCl solution absorbs 71.56 J of heat when the temperature changes from 18.6 C to 29.7 C.