Answer : The enthalpy of fusion of ice is, 334 J
Solution :
Formula used :
\(q=m\times \Delta H_{fusion}\)
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.
\(6680J=(20g)\times \Delta H_{fusion}\)
\(\Delta H_{fusion}=334J\)
Therefore, the enthalpy of fusion of ice is, 334 J
