The enthalpy change for converting 10.0 g of ice at -25.0°c to water at 80.0°c is kj. the specific heats of ice, water, and steam are 2.09 j/gk, 4.18 j/gk, and 1.84 j/gk, respectively. for h2o, δhfus = 6.01 kj/mol, and δhvap = 40.67 kj/mol.

Answer : The enthalpy change is, 7.205 KJ

Solution :

The conversions involved in this process are :

Now we have to calculate the enthalpy change.

where,

= enthalpy change = ?

m = mass of water = 10 g

= specific heat of solid water = 2.09 J/gk

= specific heat of liquid water = 4.18 J/gk

n = number of moles of water =

= enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

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

    (1 KJ = 1000 J)

Therefore, the enthalpy change is, 7.205 KJ