Calculate the final temperature of 215 mL of water initially at 40 ∘C upon absorption of 16 kJ of heat.
Assuming the density of H2O is 1 g/mL
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
16,000 = 215 x 4.184 x (Tfinal-40)
To calculate the final temperature of the water, we need to use the specific heat capacity equation:
Q = m * c * ΔT
Where:
Q = heat absorbed or released by the substance (in this case, 16 kJ)
m = mass of the substance (in this case, 215 mL of water)
c = specific heat capacity of the substance (4.18 J/g°C for water)
ΔT = change in temperature
First, we need to convert the volume of water to its mass. We can do this using the density of water, which is approximately 1 g/mL.
Mass of water = volume of water * density of water
= 215 mL * 1 g/mL
= 215 g
Next, we convert the given heat from kJ to J:
16 kJ * 1000 J/kJ = 16000 J
Now we can rearrange the equation to solve for the change in temperature, ΔT:
ΔT = Q / (m * c)
ΔT = 16000 J / (215 g * 4.18 J/g°C)
ΔT ≈ 18.43 °C
Finally, we can calculate the final temperature:
Final temperature = Initial temperature + ΔT
= 40 °C + 18.43 °C
≈ 58.43 °C
Therefore, the final temperature of the water will be approximately 58.43 °C.