Question

How much heat is required to vaporize a 3 g ice cube initially at 0◦C? The latent heat of fusion of ice is 80 cal/g and the latent heat of vaporization of water is 540 cal/g. The specific heat of water is 1 cal/(g · ◦ C). Answer in units of cal.

Answer 1

Answer : The amount of heat required is, 2160 cal.

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:

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

where,

$Q$ = heat required for the reaction = ?

m = mass of ice = 3 g

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

$\Delta H_{fusion}$ = enthalpy change for fusion = $80cal/g$

$\Delta H_{vap}$ = enthalpy change for vaporization = $540cal/g$

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

$Q=[3g\times 80cal/g]+[3g\times 1cal/g^oC\times (100-0)^oC]+[3g\times 540cal/g]$

$Q=2160cal$

Therefore, the amount of heat required is, 2160 cal.