Question

The heat of fusion for water is 80. cal/g, and the heat of vaporization of water is 540 cal/g. How many calories are required to convert 10.0 g of ice at 0°C to steam at 100°C?

Answer 1

Answer : The heat required is, 7200 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)(100^oC)\\\\(3):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)`

The expression used will be:

`\Delta H=m* \Delta H_(fusion)+[m* c_(p,l)* (T_(final)-T_(initial))]+m* \Delta H_(vap)`

where,

m = mass of ice = 10.0 g

`c_(p,l)` = specific heat of liquid water =
`1cal/g^oC`

`\Delta H_(fusion)` = enthalpy change for fusion =
`80.0cal/g`

`\Delta H_(vap)` = enthalpy change for vaporization =
`540cal/g`

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

`\Delta H=10.0g* 80.0cal/g+[10.0g* 1cal/g^oC* (100-0)^oC]+10.0g* 540cal/g`

`\Delta H=7200cal`

Therefore, the heat required is, 7200 calories.