1)
$\phantom{X}m\,=\,100\,g\phantom{X}$; $\phantom{X}Q\,=\,500\,cal\phantom{X}$
$\phantom{X}\Delta\theta\,=\,30^oC\,-\,20^oC\,=\,10^oC\phantom{X}$
2)
A capacidade térmica do corpo é dada pela fórmula:
$\phantom{X}\mathbb{C}\,=\,\dfrac{Q}{\;\Delta\theta\;}\,=\,\dfrac{\;500\,cal\;}{10^oC}\;\Rightarrow\;\boxed{\;\mathbb{C}\,=\,50\,cal/^oC\;}\phantom{X}$
3)
Para obtermos o calor específico, utilizar a fórmula:
$\phantom{X}Q\,=\,m\,c\,\Delta\theta\;\Rightarrow\;c\,=\,\dfrac{Q}{\;m\Delta\theta\;}\;\Rightarrow\phantom{X}$ $\phantom{X}c\,=\,\dfrac{500 cal}{\;100g\,\centerdot\,10^oC\;}\;\Rightarrow$ $\;\boxed{\;c\,=\,0,50\,cal/g^oC\;}\phantom{X}$
*)
Outra fórmula poderia ser:
$\phantom{X}\mathbb{C}\,=\,m\centerdot c\;\Rightarrow\;c\,=\,\dfrac{\mathbb{C}}{\;m\;}\;=$ $\;\dfrac{\;50 cal/^oC\;}{100\,g}\phantom{X}$
