Answer:
[tex]C=0.91\frac{J}{g\°C}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it is possible to recall the equation to calculate the heat, Q, in these calorimetry problems as shown below:
[tex]Q=mC(T_f-T_i)[/tex]
Thus, given the absorbed heat, mass and temperatures, we can easily calculate the specific heat of the metal as shown below:
[tex]C=\frac{Q}{m(T_f-T_i)}[/tex]
Then, by plugging in we obtain:
[tex]C=\frac{1350J}{55.0g(47.0\°C-20.0\°C)} \\\\C=0.91\frac{J}{g\°C}[/tex]
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