If 689000 J of energy are added to 4.00 L of

water at 284 K, what will the final temperature of the water be?
Answer in units of K

To find the final temperature of the water, we can use the equation:

Q = mcΔT

Where:
Q is the heat energy transferred to the water (689000 J),
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.

To calculate the mass of the water, we can use the relationship between volume (V), density (ρ), and mass (m):

m = V * ρ

The density of water is 1 g/mL or 1000 g/L.

For the specific heat capacity of water, the value is approximately 4.18 J/g·K.

We can rearrange the equation Q = mcΔT to solve for ΔT:

ΔT = Q / mc

Now we can plug in the values:

Q = 689000 J
m = V * ρ = 4.00 L * 1000 g/L = 4000 g
c = 4.18 J/g·K

ΔT = 689000 J / (4000 g * 4.18 J/g·K)

Simplifying this calculation, we get:

ΔT = 41.27 K

Finally, to find the final temperature, we need to add the change in temperature to the initial temperature (284 K):

Final temperature = 284 K + ΔT

So the final temperature of the water will be:

284 K + 41.27 K = 325.27 K

Therefore, the final temperature of the water will be approximately 325.27 K.