解：?$(1)P_{高溫}=\frac {U^2}{R_{1}}=\frac {(220\ \text {V})^2}{44 \ Ω}=1.1×10^{3}\ \text {W}$?
?$(2)I_{ }=\frac {U_{ }}{R_{1}+R_{2}}=\frac { 220\ \text {V}}{ 44 \ Ω+ 2156 \ Ω}=0.1\ \text {A}$?
?$Q=UIt= 220\ \text {V}×0.1\ \text {A}×600\ \text {s}=1.32×10^{4}\ \text {J}$?
?$(3)P_{高溫}=UI= 220\ \text {V}×0.1\ \text {A}=22\ \text {W}$?
?$R_{總}(cāng)'=\frac {U^2}{P_{高溫}×110\%}=\frac {(220\ \text {V})^2}{22\ \text {W}×110\%}=2000 \ Ω$?
?$R_{2}'=R_{總}(cāng)'-R_{1}=2000 \ \mathrm {Ω}- 44 \ \mathrm {Ω}=1956 \ \mathrm {Ω}$?
