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

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error