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

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

Step-by-step 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\,(J)/(g\cdot ^(\circ)C)`,
`T_(f) = 0\,^(\circ)C`,
`T_(o) = -50\,^(\circ)C` and
`L_(f) = 334\,(J)/(g )`, then the amount of heat is:

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

`Q = 8740\,J`

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