# How much heat (in kJ) is needed to convert 866 g of ice at -10˚C to steam at 126 ˚C? (The specific heats of ice and steam are 2.03 J/g. ˚C and 1.99 J/g. ˚C, respectively.)

May 15, 2018

2714kJ

#### Explanation:

This'll be a bit lengthy so bear with me.

This problem will require 5 values;

The heat required to raise the ice to its melting point (0°C), ${q}_{1}$

The heat required to melt all of said ice, ${q}_{2}$

The heat required to raise the temperature of the water to its boiling point (100°C), ${q}_{3}$

The heat required to vaporize all the water, ${q}_{4}$

And the heat required to raise the temperature of the steam to 126°C, ${q}_{5}$

We can find every value using $q = m c \Delta T$, $q = m {H}_{f u s i o n}$, and $q = m {H}_{\text{vaporization}}$

$q$ = Heat/energy
$m$ = Mass
$c$ = Specific heat
$\Delta T$ = Change in temperature
${H}_{f u s i o n}$ = Energy required to melt/freeze one gram of a substance
${H}_{\text{vaporization}}$ = Energy required to vaporize/condense one gram of a substance

I'll now write out every equation with the given info.

q_1 = 866g * (2.03J)/(g°C) * 10°C

${q}_{2} = 866 g \cdot \frac{334 J}{g}$

q_3 = 866g * (4.18J)/(g°C) * 100°C

${q}_{4} = 866 g \cdot \frac{2260 J}{g}$

q_5 = 866g * (1.99J)/(g°C) * 26°C

Note: in the 3rd and 4 equations, 334J/g and 2260J/g are the heat of fusion and vaporization of water respectively.

After calculating and using sig figs in each equation, we are given 5 values:

${q}_{1} = 20000 J$
${q}_{2} = 289000 J$
${q}_{3} = 400000 J$
${q}_{4} = 1960000 J$
${q}_{5} = 45000 J$

After adding all the values, we have a total of 2714000J, or 2714kJ required for the conversion.