What mass of steam is required to release
3.15×105
kJ of heat energy on condensation?
Answer in units of g.
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To calculate the mass of steam required to release a certain amount of heat energy on condensation, we need to use the equation:
Q = m * ΔH
Where:
Q is the heat energy released (given as 3.15×10^5 kJ)
m is the mass of steam
ΔH is the enthalpy of condensation of steam (specific heat of vaporization) in kJ/g
To find the value of ΔH, we can refer to a steam table or look it up online. The enthalpy of condensation of steam is approximately 2257 kJ/kg or 2.257 kJ/g.
Substituting the values into the equation, we have:
3.15×10^5 kJ = m * 2.257 kJ/g
Now, let's solve for the mass of steam (m):
m = (3.15×10^5 kJ) / (2.257 kJ/g)
m ≈ 139588.62 g
Therefore, approximately 139588.62 grams (or 139.6 kg) of steam is required to release 3.15×10^5 kJ of heat energy on condensation.