Question

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. cal/g.

Answer 1

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.