When 19 mL of water at 22 C is combined with 18 mL of water at 64 C, what is the final temperature of the water?. The specific heat of water is 4.184 J/g/K and its density is 1.0 g/mL.

c_hot = (4.184J/g*K)(18mL)(1g/mL)

c_cold = (4.184J/g*K)(19mL)(1g/mL)
Use:
0 = c_hot(T_f - T_initial,hot)+ c_cold(T_f - T_initial,cold)
Solve for T_f and remember to convert temperature back to C from Kelvin and vice versa.

Best of luck,
ChemVantage Loather

Just a quick note to say that changing celsius to kelvin and back again is not necessary. Since it is a difference of temperatures only, and since 1 degree C = 1 degree K, you may subtract with C or K and come up with the same value.

To find the final temperature of the water, we can use the principle of conservation of energy. The heat gained by the cooler water is equal to the heat lost by the hotter water.

First, let's calculate the heat lost by the hotter water. We can use the formula:

Q = mcΔT

Where:
Q is the heat,
m is the mass of the water,
c is the specific heat of water, and
ΔT is the change in temperature.

Since the density of water is 1.0 g/mL, we know that the mass of the water is equal to its volume. Therefore, for the hotter water:

m1 = 18 mL = 18 g

Now, let's calculate the heat gained by the cooler water. Similarly:

m2 = 19 mL = 19 g

Next, let's calculate the change in temperature for both the hotter and cooler water. We'll call it ΔT1 and ΔT2 respectively.

ΔT1 = final temperature - initial temperature
ΔT2 = final temperature - initial temperature

Using the given temperatures:
Initial temperature for the hotter water = 64°C
Initial temperature for the colder water = 22°C

Now, let's plug the values into the formulas and set the heat gained equal to the heat lost:

m1c(64 - final temperature) = m2c(final temperature - 22)

Simplifying the equation:

18 * 4.184 * (64 - final temperature) = 19 * 4.184 * (final temperature - 22)

Now, we can solve for the final temperature.