## chris215 one year ago -

1. Michele_Laino

we have to compute the amount Q_1 of heat needed to melt that quantity of ice, and the amount Q_2 of heat needed to raise the temperature of the water from zero to 100 °celsius degree, and the amount Q_3 of heat to transform all the water vapor in steam at 100 degrees. Now we have this: $\Large \begin{gathered} {Q_1} = {\lambda _F}m = 79.7 \times 2 = ...kCal \hfill \\ \hfill \\ {Q_2} = {c_{{H_2}O}}m\Delta t = 1 \times 2 \times 100 = ...kCal \hfill \\ \hfill \\ {Q_3} = {\lambda _E}m = 539.2 \times 2 = ...kCal \hfill \\ \end{gathered}$ where: $\Large {\lambda _F},\;{\lambda _E}$ are the latent heats of melting and vaporization of water respectively, and $\Large {c_{{H_2}O}}$ is the specific heat of water Finally, the requested amount Q_Total of heat is given by the subsequent formula: $\Large {Q_{Total}} = {Q_1} + {Q_2} + {Q_3} = ...$ Please complete my computation

2. Michele_Laino

that's right!