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You add 100.0 g of water at 52.0 °C to 100.0 g of ice at 0.00 °C. Some of the ice melts and cools the water to 0.00 °C. When the ice and water mixture reaches thermal equilibrium at 0 °C, how much ice has melted? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K. The enthalpy of fusion of ice at 0 °C is 333 J/g.) Mass of ice = g

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Answer:

[tex]m_{ice} = 65.336\,g[/tex]

Explanation:

Accoding to the First Law of Thermodynamics, the heat released by the water melts a portion of ice. That is to say:

[tex]Q_{water} = Q_{ice}[/tex]

[tex](100\,g)\cdot \left(4.184\,\frac{J}{kg\cdot ^{\textdegree}C}\right)\cdot (52\,^{\textdegree}C - 0\,^{\textdegree}C) = m_{ice}\cdot \left(333\,\frac{J}{g} \right)[/tex]

The amount of ice that is melt is:

[tex]m_{ice} = 65.336\,g[/tex]

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