Answer
a) The answer to this question can be found in the attached picture.
b) The answer is $k=\frac{1}{273^{\circ}C}.$
c) The answer is $R_{0}\approx 1.02 \Omega$.
d) The answer is $T\approx 128.5 ^{\circ} C$.
Work Step by Step
a) The steps are included in the attached picture.
b) The resistance is equal to zero $R=0\Omega$ when the temperature is $T=-273^{\circ}C$. Thus, we plug these values to the equation to acquire
$R=R_{0}(1+k T)\Rightarrow$$0=R_{0}[1+k (-273)]$. Since $R_{0}>0$ we can divide both sides to acquire $1+k(-273^{\circ}C)=0\Rightarrow$$k(-273^{\circ}C)=-1\Rightarrow$$k=\frac{1}{273^{\circ}C}.$
c) For a tungsten filament it is given $R=1.1\Omega$ when $T=20^{\circ}C$. Given that $k$ is constant for all pure metal wires, then we can use the value obtained in the previous question. To this end,
$R=R_{0}(1+k T)\Rightarrow$$1.1\Omega=R_{0}(1+\frac{1}{273^{\circ}C}20^{\circ}C)\Rightarrow$$1.1\Omega=R_{0}(\frac{273+20}{273})\Rightarrow$$1.1\Omega=R_{0}\frac{293}{273}\Rightarrow$$R_{0}=\frac{1.1*273}{293}\Omega\Rightarrow$$R_{0}=1.02\Omega$.
d) Given the values of $R_{0}, k$ from the previous questions, now we seek the value of T when $R=1.5\Omega$. Once more
$R=R_{0}(1+k T)\Rightarrow$$1.5\Omega=1.02\Omega(1+\frac{1}{273^{\circ}C}T)\Rightarrow$$\frac{1.5\Omega}{1.02\Omega}=1+\frac{T}{273^{\circ}C}\Rightarrow$$1.47-1=\frac{T}{273^{\circ}C}\Rightarrow$$T=273*0.47^{\circ}C\Rightarrow$$T\approx 128.5^{\circ}C.$
