Answer
a) $253^\circ\rm C$
b) $32.6\;\rm cm$
Work Step by Step
a) The gas is heated under constant pressure which means that it is an isobaric process. So the heat transferred to gas is given by
$$Q=nC_{\rm p}\Delta T$$
$$Q=nC_{\rm p}(T_f-T_i) $$
Hence, the final temperature is given by
$$T_f=\dfrac{Q}{nC_{\rm p}}+T_i\tag 1$$
And since Argon is considered an ideal gas, the number of moles is given by
$$P_iV_i=nRT_i$$
$$n=\dfrac{P_iV_i}{RT_i}$$
The initial volume is given by $V_i=AL_i=\pi r^2L_i$
$$n=\dfrac{\pi r^2 L_i P_i }{RT_i}$$
Plugging into (1);
$$T_f=\dfrac{QRT_i}{ \pi r^2 L_i P_i C_{\rm p}}+T_i $$
Plugging the known;
$$T_f=\dfrac{(2500)(8.31)(50+273)}{ \pi (0.05)^2 (0.20)(10\times 1.013\times 10^5)(20.8)}+(50+273)$$
$$T_f=526\;\rm K\approx \color{red}{\bf 253^\circ} \rm C$$
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b) To find the final length, we need to solve for the final volume.
We know that the gas undergoes an isobaric process, so
$$\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f}$$
$$\dfrac{\color{red}{\bf\not} AL_i}{T_i}=\dfrac{\color{red}{\bf\not} AL_f}{T_f}$$
Hence,
$$L_f=\dfrac{L_iT_f}{T_i}$$
Plugging the known;
$$L_f=\dfrac{(20)(526)}{(50+273)}$$
$$L_f =\color{red}{\bf 32.6 } \rm \;cm$$
