Respuesta :
ANSWER
[tex]18.05\text{ }kcal[/tex]EXPLANATION
The amount of heat necessary to change the ice to water is given by:
[tex]Q=m(c_{ice}\Delta T_{ice}+L+c_{water}\Delta T_{water})[/tex]where m = mass of ice = 480 g = 0.48 kg
c(ice) = specific heat capacity of ice = 2108 J/kg/K
ΔTice = change in temperature of ice = 0 - (-19) = 19 K or 19 °C
L = latent heat of fusion of ice = 33600 J/k
c(water) = specific heat capacity of water = 4186 J/kg/K
ΔTwater = change in temperature of water = 20 - 0 = 20 K or 20 °C
Therefore, the heat necessary is:
[tex]\begin{gathered} Q=0.48([2108*19]+33600+[4186*20]) \\ \\ Q=0.48(40052+33600+83720)=0.48*157372 \\ \\ Q=75538.56\text{ }J \end{gathered}[/tex]Convert this to kcal:
[tex]\begin{gathered} 1\text{ }J=\frac{1}{4184}\text{ }kcal \\ \\ 75538.56\text{ }J=\frac{75538.56}{4184}\text{ }kCal=18.05\text{ }kcal \end{gathered}[/tex]That is the answer.
