Question

How much heat is required to convert 20.0 g of ice at 50.0⁰C to liquid water at 0.0⁰C? The specific heat of ice is 2.06 J/(g∙⁰C) and the heat of fusion of water is 334 J/g.

Answer 1

Answer:

8740 joules are required to convert 20 grams of ice to liquid water.

Explanation:

The amount of heat required ($Q$), 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:

$Q = m\cdot [c\cdot (T_{f}-T_{o})+L_{f}]$ (1)

Where:

$m$ - Mass, measured in grams.

$c$ - Specific heat of ice, measured in joules per gram-degree Celsius.

$T_{o}$, $T_{f}$ - Temperature, measured in degrees Celsius.

$L_{f}$ - Latent heat of fussion, measured in joules per gram.

If we know that $m = 20\,g$, $c = 2.06\,\frac{J}{g\cdot ^{\circ}C}$, $T_{f} = 0\,^{\circ}C$, $T_{o} = -50\,^{\circ}C$ and $L_{f} = 334\,\frac{J}{g }$, then the amount of heat is:

$Q = (20\,g)\cdot \left\{\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot [0\,^{\circ}C-(-50\,^{\circ}C)]+334\,\frac{J}{g} \right\}$

$Q = 8740\,J$

8740 joules are required to convert 20 grams of ice to liquid water.