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Gwar [14]
3 years ago
11

How many calories are required to convert 17 g of ice at 0.0°C to liquid water at 32.0°C? The heat of fusion of water is 80. c

al/g.
Chemistry
1 answer:
Mazyrski [523]3 years ago
3 0

Answer : The heat required is, 1904 calories.

Explanation :

The process involved in this problem are :

(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(32.0^oC)

The expression used will be:

\Delta H=m\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]

where,

m = mass of ice = 17 g

c_{p,l} = specific heat of liquid water = 1cal/g^oC

\Delta H_{fusion} = enthalpy change for fusion = 80.0cal/g

Now put all the given values in the above expression, we get:

\Delta H=17g\times 80.0cal/g+[17g\times 1cal/g^oC\times (32.0-0)^oC]

\Delta H=1904cal

Therefore, the heat required is, 1904 calories.

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