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:

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

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:

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

$\Delta H=7200cal$

Therefore, the heat required is, 7200 calories.