The enthalpy change for converting 10.0 g of ice at -25.0 °c to water at 90.0 °c is kj. the specific heats of ice, water, and steam are 2.09 j/g-k, 4.18 j/g-k, and 1.84 j/g-k, respectively. for h2o, = 6.01 kj/mol, and = 40.67 kj/mol.

Answer : The enthalpy change is, 6.2315 KJ

Solution :

The conversions involved in this process are :

Now we have to calculate the enthalpy change.

where,

= enthalpy change = ?

m = mass of ice = 10.0 g

= specific heat of solid water =

= specific heat of liquid water =

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, 6.2315 KJ