解:$(1)Q_吸=cm∆t=4.2×10^3\ \text{J/(kg}·℃) ×{ 1 }\ \text{kg}×(100℃-20℃)={ 3.36×10^5 }\ \text{J}$
$ (2)W=Pt={ 1100 }\ \text {W}×{ 6×60 }\ \text {s}={ 3.96×10^5 }\ \text {J}$
$ η=\frac {Q_吸}{W}×100\%=\frac {{ 3.36×10^5 }\ \text{J}}{{ 3.96×10^5 }\ \text{J}}×100\%≈{ 84.8 }\%$
$ (3)R_{ }=\frac {U_{ 额 }^2}{P_{额 }}=\frac {({ 220 }\ \text {V})^2}{{ 1100 }\ \text {W}}={ 44 }Ω$
$ P_{ 实 }=\frac {U_{实 }^2}{R_{ }}=\frac {({ 198 }\ \text {V})^2}{{ 44}Ω}={ 891}\ \text {W}$