Answer:
![47.07g](https://img.qammunity.org/2021/formulas/physics/college/afqdtfej0x7h1rdf3stlzyalu8cbzzgwji.png)
Step-by-step explanation:
500 watts is equivalent to 500J/s
20 minutes is equivalent to 20*60seconds which is 1200 s
330g is equivalent to 330/1000kg which is 0.33kg
The temperature will be (100-23) which is 77°C
The energy will be:
![Energy=500J/s * 1200s](https://img.qammunity.org/2021/formulas/physics/college/ku9ebsaoq02ex5e6na7do7yxed5ov014mw.png)
![Energy=600kJ](https://img.qammunity.org/2021/formulas/physics/college/mql87ezy4ceozmz87obcomd70xnpkb4xe3.png)
Now we can apply:
Q=
*m*ΔT
Q=4.186kJ/kgC * 0.33kg * 77°C
Q=106.37kJ
Energy to boil water is 2260J/kg
2260J/kg*0.33=
so The energy available will be:
=745.8kJ-106.37kJ
=639.43kJ
Now the remaining water will be:
=(1-639.43kJ/745.8kJ)*0.33
![=47.07g](https://img.qammunity.org/2021/formulas/physics/college/l785v79uvqvzr1jw6rlbw2zz419fyvcf8f.png)