Calculate the enthalpy change (in kj) associated with the conversion of 25.0 grams of ice at -4.00 °c to water vapor at 108.0 °c. 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, δ hfus

Q = 75652 J

Explanation:

as we know that we have 25 g ice at -4 degree C

so here we can say

heat require to convert the ice into steam at 108 degree is in following steps

1) First we need to convert the ice into 0 degree C then it has to convert into water at 0 degree C

so we have

2)Now water at 0 degree C is to be converted into water at 100 degree C

3) Now water at 100 degree C is to be converted into steam and then raise the temperature from 100 degree C to 108 degree C

now we have

Now total heat required to raise the temperature of ice is given as

explanation:

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