Answer:
[tex]C_{metal}=126.6\frac{J}{g\°C}[/tex]
Explanation:
Hello!
In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:
[tex]Q_{metal}=-Q_{water}[/tex]
Which can be also written as:
[tex]m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})[/tex]
Thus, since we need the specific heat of the metal, we solve for it as shown below:
[tex]C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}[/tex]
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