Answer:
[tex]C=0.152\frac{J}{g\°C}[/tex]
Explanation:
Hello!
In this case, when doing calorimetry problems, we need to use the following equation:
[tex]Q=mC(T_2-T_1)[/tex]
Whereas Q is the involved heat, m the mass, C the specific heat and T2 and T1 the final and initial temperatures respectively. Thus, since we need the specific heat, we proceed as follows:
[tex]C=\frac{Q}{m(T_2-T_1)}[/tex]
Thus, we plug in to obtain:
[tex]C=\frac{15,650J}{750g(250\°C-25\°C)} \\\\C=0.152\frac{J}{g\°C}[/tex]
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