$解：(1)I_{ }=\frac {U_{ R}}{R_{ }}=\frac {{ 2 }\ \text {V}}{{ 20 }Ω}={ 0.1 }\ \text {A}$
$U_t=IR_t=0.1\ \text {A}×40Ω=4\ \text {V}$
$U=U_R+U_t=2\ \text {V}+4\ \text {V}=6\ \text {V}$
$(2)I_{R }=\frac {U_{ }}{R_{ }}=\frac {{ 6 }\ \text {V}}{{ 20 }Ω}={ 0.3 }\ \text {A}$
$I'_t=I'-I_R=0.5\ \text {A}-0.3\ \text {A}=0.2\ \text {A}$
$R_{ t }'=\frac {U_{ }}{I_{ t }'}=\frac {{ 6 }\ \text {V}}{{ 0.2 }\ \text {A}}={ 30 }Ω$
$由图乙知，此时温度为30℃.$
