The heat of solution of potassium acetate in water is -15.3 ,What will be the final temperature when 0.132 is dissolved in 550 water that is initially at 25.1 ?

You get 15.3 WHAT? for 1 mol KC2H3O2 (98.1g)

15.3? x (0.132/98.1) = q
Then q = mass H2O x specific heat H2O x (Tfinal-Tinitial)

To determine the final temperature when 0.132 g of potassium acetate is dissolved in 550 g of water initially at 25.1 °C, we can use the equation:

q = m * c * ΔT

Where:
q = heat transferred (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity (in J/g °C)
ΔT = change in temperature (in °C)

First, let's calculate the heat transferred when the potassium acetate is dissolved:

q1 = m1 * ΔH

Where:
q1 = heat transfer during dissolution (in Joules)
m1 = mass of the potassium acetate (in grams)
ΔH = heat of solution (in J/g)

Given:
m1 = 0.132 g
ΔH = -15.3 J/g

q1 = 0.132 g * (-15.3 J/g)
q1 = -2.0196 J

Next, let's calculate the heat transferred when the water is heated up:

q2 = m2 * c2 * ΔT2

Where:
q2 = heat transfer during heating (in Joules)
m2 = mass of the water (in grams)
c2 = specific heat capacity of water (in J/g °C)
ΔT2 = change in temperature of water (in °C)

Given:
m2 = 550 g
c2 = 4.18 J/g °C
ΔT2 = final temperature - initial temperature

Let's rearrange the equation to solve for ΔT2:

q2 = m2 * c2 * ΔT2
ΔT2 = q2 / (m2 * c2)

We need to find q2, so let's use the fact that the heat transferred in an exothermic dissolution is equal to the heat gained by the water:

q2 = -q1
q2 = -(-2.0196 J)
q2 = 2.0196 J

Now, let's calculate ΔT2:

ΔT2 = 2.0196 J / (550 g * 4.18 J/g °C)
ΔT2 ≈ 0.000867 °C

To find the final temperature, we need to add ΔT2 to the initial temperature:

Final Temperature = 25.1 °C + 0.000867 °C
Final Temperature ≈ 25.1009 °C

Therefore, the final temperature when 0.132 g of potassium acetate is dissolved in 550 g of water initially at 25.1 °C is approximately 25.1009 °C.

To find the final temperature when potassium acetate is dissolved in water, we can use the equation:

q = m × c × ΔT

where:
- q is the heat gained or lost by the system (water and potassium acetate)
- m is the mass of the water in grams
- c is the specific heat of water (which is approximately 4.18 J/g°C)
- ΔT is the change in temperature in degrees Celsius

Since we are given the heat of solution (q) and the mass of water (m), we can rearrange the equation to solve for ΔT:

ΔT = q / (m × c)

Now let's substitute the given values into the equation:

Mass of water (m) = 550 g
Heat of solution (q) = -15.3 kJ (Note: we need to convert this to joules)
Specific heat of water (c) = 4.18 J/g°C

First, let's convert the heat of solution from kJ to joules:

-15.3 kJ × 1000 J/1 kJ = -15,300 J

Now we can substitute the values:

ΔT = -15,300 J / (550 g × 4.18 J/g°C)

Calculating the value:

ΔT = -28.91 °C

Since the initial temperature is 25.1 °C, the final temperature will be:

Final temperature = Initial temperature + ΔT
Final temperature = 25.1 °C + (-28.91 °C)
Final temperature ≈ -3.81 °C

Therefore, the final temperature when 0.132 g of potassium acetate is dissolved in 550 g of water that is initially at 25.1 °C will be approximately -3.81 °C.