1)
Dados do enunciado:
$\phantom{X}Q\,=\,1\,BTU\phantom{X}$;$\phantom{X}m\,=\,1\,\ell b\,\cong\,454\,g\phantom{X}$
como $\,\dfrac{\;\Delta\theta\;C\;}{\;\Delta\theta\;F\;}\,=\,\dfrac{\;100\;}{\;180\;}\,=\,\dfrac{\;5\;}{9}\;$ então
$\phantom{X}\Delta\theta\,=\,65^oF\,-\,64^oF\,=\,1^oF\,=\,\dfrac{\;5\;}{9}\;{}^oC\phantom{X}$
2)
O calor específico da água é cH2O = 1 cal/g°C e usando a fórmula Q = mcΔθ
$\phantom{X}1\;BTU\,=\,454g\,\centerdot\,1\,\dfrac{cal}{g\,\centerdot\,{}^oC}\,\centerdot\,\dfrac{5}{9}{}^oC\;\Rightarrow$ $\;\boxed{\;1\,BTU\,\cong\,252\,cal\;}\phantom{X}$
