Find the mass of liquid water required to
absorb 4.73×104
kJ of heat energy on boiling.
Answer in units of g.
To find the mass of liquid water required to absorb 4.73×10^4 kJ of heat energy on boiling, we can use the equation:
q = m × ΔH
Where:
q is the amount of heat absorbed or released
m is the mass of the substance
ΔH is the specific heat of the substance
In this case, we need to find the mass, so we rearrange the equation to solve for m:
m = q / ΔH
The specific heat of water is 4.18 J/g°C, which is equal to 4.18 kJ/g°C (since 1 kJ = 1000 J and the units cancel out).
Now, we can substitute the given values into the equation and calculate the mass:
m = 4.73×10^4 kJ / 4.18 kJ/g°C
m ≈ 11315.07 g
Therefore, the mass of liquid water required to absorb 4.73×10^4 kJ of heat energy on boiling is approximately 11315.07 grams (g).
q = mass H2O x dHvap = ?
Watch the units.