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Answer:
8740 joules are required to convert 20 grams of ice to liquid water.
Explanation:
The amount of heat required ([tex]Q[/tex]), measured in joules, to convert ice at -50.0 ºC to liquid water at 0.0 ºC is the sum of sensible heat associated with ice and latent heat of fussion. That is:
[tex]Q = m\cdot [c\cdot (T_{f}-T_{o})+L_{f}][/tex] (1)
Where:
[tex]m[/tex] - Mass, measured in grams.
[tex]c[/tex] - Specific heat of ice, measured in joules per gram-degree Celsius.
[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Temperature, measured in degrees Celsius.
[tex]L_{f}[/tex] - Latent heat of fussion, measured in joules per gram.
If we know that [tex]m = 20\,g[/tex], [tex]c = 2.06\,\frac{J}{g\cdot ^{\circ}C}[/tex], [tex]T_{f} = 0\,^{\circ}C[/tex], [tex]T_{o} = -50\,^{\circ}C[/tex] and [tex]L_{f} = 334\,\frac{J}{g }[/tex], then the amount of heat is:
[tex]Q = (20\,g)\cdot \left\{\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot [0\,^{\circ}C-(-50\,^{\circ}C)]+334\,\frac{J}{g} \right\}[/tex]
[tex]Q = 8740\,J[/tex]
8740 joules are required to convert 20 grams of ice to liquid water.