Respuesta :
Given:
The mass of ice is,
[tex]m=460\text{ kg}[/tex]The initial temperature of ice is,
[tex]t_i=-18^{\circ}C[/tex]The heat for this ice to become 0 degree C ice is,
[tex]\begin{gathered} H_1=460\times0.5\times18 \\ =4.14\text{ kcal} \end{gathered}[/tex]The heat needed for 0 degree ice to 0 degree water is,
[tex]\begin{gathered} H_2=460\times80 \\ =36.8\text{ kcal} \end{gathered}[/tex]Now the heat for the final step to reach 20 degrees is,
[tex]\begin{gathered} H_3=460\times1\times(20-0) \\ =9.2\text{ kcal} \end{gathered}[/tex]The total heat is,
[tex]\begin{gathered} H=4.14+36.8+9.2 \\ =50.1\text{ kcal} \end{gathered}[/tex]Hence the total heat is 50.1 kcal.
