Answer:
Q = 160.36[kJ], is the heat lost.
Explanation:
This is a thermodynamic problem where we can find the latent heat of vaporization, at a constant temperature of 100 [°C].
We know that for steam at 100[C], the enthalpy is
[tex]h_{gas} = 2675.6[kJ/kg][/tex]
For liquid water at 100[C], the enthalpy is
[tex]h_{water} = 419.17[kJ/kg][/tex]
Therefore
[tex]h_{g-w} = 2675.6-419.17 = 2256.43[\frac{kJ}{kg} ][/tex]
The amout of heat is given by:
[tex]Q=h_{g-w}*m\\ where:\\m = mass = 0.07107[kg] = 71.01[g]\\Q = heat [kJ]\\Q =2256.43*0.07107\\Q=160.36[kJ][/tex]
