Answer
See the detailed answer below.
Work Step by Step
a) We need to know the two materials first so we can write a realistic question.
It seems that the $13600\;\rm kg/m^3$ is the density of mercury and that the $449\;\rm J/Kg\cdot K$ is the specific heat of Iron.
Now we can state the question:
Find the initial temperature of a 0,5-kg iron piece if we added it to 200 cm$^3$ of mercury at 15$^\circ$C when the final temperature of the mixture is 90$^\circ$C
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b) Solving for $T_i$;
$$(90+273)-T_i=-\dfrac{(200\times 10^{-6})(13,600)(140)(90-15)}{(0.5)(449)}$$
$$T_i=363+\dfrac{(200\times 10^{-6})(13,600)(140)(90-15)}{(0.5)(449)}=\bf 490\;\rm K$$
$$T_i=\color{red}{\bf 217}^\circ\rm C$$
