Dados:
[tex]m=100\ g,\ T_0=-18\ºC,\ T_1=0\ºC,\ T_2=42.6\ºC\\ c_l=5\ cal/g\ºF,\ c_s=2\ cal/g\ºF,\ L_f=90\ cal/g[/tex]
Primeiramente, devemos converter as temperaturas da escala Celsius para a escala Fahrenheit:
[tex]\boxed{\dfrac{T_C}{5}=\dfrac{T_F-32}{9}=\dfrac{T_K-273}{5}}\ \therefore\ T_F=1.8T_C+32[/tex]
[tex]T_0=-18\º C\ \therefore\ T_0=-0.4\º F\\\\ T_1=0\ºC\ \therefore\ T_1=32\ºF\\\\ T_2=42.6\º\ \therefore\ T_2=108.68\ºF[/tex]
Dessa forma, podemos calcular o calor liberado:
[tex]Q=Q_{0\to1}+Q_{1\to2}+Q_f\ \therefore\\\\ Q=mc_s(T_1-T_0)+mc_l(T_2-T_1)+mL_f\ \therefore\\\\ Q=m\bigg[c_s(T_1-T_0)+c_l(T_2-T_1)+L_f\bigg]\ \therefore[/tex]
[tex]Q=100\bigg[2(32-(-0.4))+5(108.68-32)+90\bigg]\ \therefore[/tex]
[tex]Q=53820\ cal\ \therefore\ \boxed{Q\approx54\ kcal}[/tex]
Letra E.
