Answer:
The specific heat of the metal is [tex]0.337\ J/g^{\circ} C[/tex].
Explanation:
We have,
Mass of sample is 22.44 g
It absorbs 180.8 J of heat.
The temperature of the sample increases from 21.1 °C to 45.0 °C
Initial temperature is 21.1 °C and final is 45.0 °C.
Heat absorbed in terms of specific heat is given by :
[tex]Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{180.8}{22.44\times (45-21.1)}\\\\c=0.337\ J/g^{\circ} C[/tex]
So, the specific heat of the metal is [tex]0.337\ J/g^{\circ} C[/tex].
