Question

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

Answer 1

Answer:

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

$Q_1 = ms\Delta T_1 + mL$

$Q_1 = 25(2.09)(0 -(-4)) + 25(335)$

$Q_1 = 8584 J$

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

$Q_2 = ms\Delta T_2$

$Q_2 = 25(4.18)(100 - 0)$

$Q_2 = 10450$

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

$Q_3 = ms\Delta T_3 + mL$

$Q_3 = 25(1.84)(108 - 100) + 25(2250)$

$Q_3 = 56618 J$

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

$Q = Q_1 + Q_2 + Q_3$

$Q = 8584 + 10450 + 56618 = 75652 J$