If 765000 J of energy are added to 7.90 L of
water at 291 K, what will the final temperature
of the water be?
Answer in units of K.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Substitute and solve for Tf.
I think you use the formula q=mc(delta)t
with q being the amount of energy added, m being the mass of water,c being the specific heat capacity of water which is 4.18J/g*degrees Celsius and then you would solve for delta t which is your change in temperature... hope this helps somewhat :)
To find the final temperature of the water, we can use the formula:
\(q = mcΔT\)
Where:
\(q\) is the heat energy gained or lost by the substance,
\(m\) is the mass of the substance,
\(c\) is the specific heat capacity of the substance,
and \(\Delta T\) is the change in temperature.
First, we need to calculate the mass of the water. We know that the volume is 7.90 L, and the density of water is approximately 1 g/mL or 1000 kg/m^3. Therefore, the mass of the water is:
\(m = V \times ρ = 7.90 \, \text{L} \times 1000 \, \text{kg/m}^3 = 7900 \, \text{g}\)
Next, we need to determine the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/(g·K).
Now, we'll rearrange the formula to solve for the change in temperature:
\(ΔT = \frac{q}{mc}\)
Substituting the given values, we have:
\(ΔT = \frac{765000 \, \text{J}}{7900 \, \text{g} \times 4.18 \, \text{J/(g·K)}}\)
Calculating this value, we get:
\(ΔT ≈ 23.02 \, \text{K}\)
Finally, to find the final temperature, we add the change in temperature to the initial temperature (291 K):
\(T_{\text{final}} = T_{\text{initial}} + ΔT = 291 \, \text{K} + 23.02 \, \text{K}\)
Calculating this expression, we find:
\(T_{\text{final}} ≈ 314.02 \, \text{K}\)
Therefore, the final temperature of the water will be approximately 314.02 K.