Answer:
Q = 160.36[kJ], is the heat lost.
Step-by-step 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
![h_(gas) = 2675.6[kJ/kg]](https://img.qammunity.org/2021/formulas/physics/middle-school/j7i0gvsr7vo3m4ywuhkzfg1y7w5cnrzvb5.png)
For liquid water at 100[C], the enthalpy is
![h_(water) = 419.17[kJ/kg]](https://img.qammunity.org/2021/formulas/physics/middle-school/cqwcb0obvcgxaeityy2fwrd1vfgn6waqeu.png)
Therefore
![h_(g-w) = 2675.6-419.17 = 2256.43[(kJ)/(kg) ]](https://img.qammunity.org/2021/formulas/physics/middle-school/welbs7u3791cvdaxpnr3ib35c52j9goc17.png)
The amout of heat is given by:
![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]](https://img.qammunity.org/2021/formulas/physics/middle-school/upd6yc4txr7zkatj75olrt7wf2h2nkxrhn.png)