1 Answer

Yulisa · Answer 1 · 2017-04-13T11:24:33+0000

Answer : The enthalpy of fusion of ice is, 334 J

Solution :

Formula used :

$q=m* \Delta H_(fusion)$

Where,

q = heat absorb = 6.68 KJ = 6680 J

m = mass of ice = 20 g

$\Delta H_(fusion)$ = enthalpy of fusion of ice = ?

Now put all the given values in this formula, we get the enthalpy of fusion of ice.

$6680J=(20g)* \Delta H_(fusion)$

$\Delta H_(fusion)=334J$

Therefore, the enthalpy of fusion of ice is, 334 J

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