Question

In a calorimeter, 6.68 kJ of heat was absorbed by 20 g of ice. What is the enthalpy of fusion of the ice? q=m∆ahf

Answer 1

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