Answer:
c=0.2 J/g°C
Explanation:
Given that,
Mass, m = 41.81 g
Initial temperature, [tex]T_i=20.2 ^{\circ} C[/tex]
Final temperature, [tex]T_f=184.6^{\circ} C[/tex]
The heat absorbed by the substance is 1375 J.
We need to find the specific heat (c) of the substance. It can be given by the below formula:
[tex]Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1375\ J}{41.81\ g\times (184.6-20.2)^{\circ} C}\\\\c=0.2\ J/g^{\circ} C[/tex]
Hence, the specific heat (c) of the substance is 0.2 J/g°C.
