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o-na [289]
3 years ago
8

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 late

nt heat of vaporization of water is 540 cal/g. The specific heat of water is 1 cal/(g · ◦ C). Answer in units of cal.
Chemistry
1 answer:
kupik [55]3 years ago
7 0

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.

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