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chris215

  • one year ago

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  1. Michele_Laino
    • one year ago
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    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
    • one year ago
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    that's right!

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