A sample of KOH is dissolved into 100.0 grams of water in a coffee cup calorimeter. if the dissolution produces 6.45 kJ of heat and the temperature change is 25.4 degrees celcius to 40.0 degrees celcius, how much KOH (in grams) is added to the water? (the specific heat of the solution is 4.20 J/g degree celcius.)

q = mass x specific heat x (Tfinal-Tinitial)

q = 6450 joules.
mass = only unknown
specific heat = 4.20 J/g
Tfinal = 40.0
Tinitial= 25.4
Note the correct spelling of celsius.

To calculate the amount of KOH (in grams) added to the water, we need to use the equation for heat transfer:

q = m * c * ΔT

where:
q = heat transferred (in J)
m = mass of the KOH (in grams)
c = specific heat of the solution (in J/g °C)
ΔT = change in temperature (in °C)

First, we need to convert the amount of heat transferred from kJ to J:

6.45 kJ * 1000 J/1 kJ = 6450 J

Next, we substitute the given values into the equation:

6450 J = m * 4.20 J/g °C * (40.0 °C - 25.4 °C)

Now we can solve for the mass, m:

6450 J = m * 4.20 J/g °C * 14.6 °C

Dividing both sides of the equation by 4.20 J/g °C * 14.6 °C:

6450 J / (4.20 J/g °C * 14.6 °C) = m

m ≈ 263.92 g

Therefore, approximately 263.92 grams of KOH were added to the water.