Answer:
Approximately [tex]41.2\; {\rm ^{\circ} C}[/tex].
Explanation:
Let [tex]t\; {\rm ^{\circ} C}[/tex] be the final temperature of the water and the iron.
Temperature of the water would be increase by [tex](t - 38.00)\; {\rm ^{\circ} C}[/tex].
Temperature of the iron would be reduced by [tex](100.0 - t)\; {\rm ^{\circ} C}[/tex].
Let [tex]c[/tex] denote the specific heat of each material. Let [tex]m[/tex] denote the mass of the material. For a temperature change of [tex]\Delta t[/tex], the energy change involved would be:
[tex]Q = c\, m \, \Delta t[/tex].
The energy that the water need to absorb would be:
[tex]\begin{aligned}& Q(\text{water, absorbed}) \\ =\; & c(\text{water}) \, m(\text{water})\, \Delta t (\text{water}) \\ =\; & 4.181\; {\rm J \cdot g^{-1} \cdot K^{-1}} \times 50\; {\rm g} \times (t - 38.00)\; {\rm ^{\circ} C} \\ =\; & (209.05\, t - 7943.9)\; {\rm J} \end{aligned}[/tex].
The energy that the iron would need to release would be:
[tex]\begin{aligned}& Q(\text{iron, released}) \\ =\; & c(\text{iron}) \, m(\text{iron})\, \Delta t (\text{iron}) \\ =\; & 0.45\; {\rm J \cdot g^{-1} \cdot K^{-1}} \times 25.0\; {\rm g} \times (100.0 - t)\; {\rm ^{\circ} C} \\ =\; & (1125 - 11.25 \, t)\; {\rm J} \end{aligned}[/tex].
Since this calorimeter is insulated, the energy that the iron had released would be equal to the energy that the water had absorbed:
[tex]Q(\text{water, absorbed}) = Q(\text{iron, released})[/tex].
[tex]209.05\, t - 7943.9 = 1125 - 11.25\, t[/tex].
[tex]t \approx 41.2[/tex].
Thus, the final temperature of the water and the iron would be approximately [tex]41.2\; {\rm ^{\circ} C}[/tex].
