# Question #2a6cb

Apr 27, 2015

You have to do 5 different calculations for this, and you need to know 5 different thermodynamic constants.

1st - heat ice from $- {5}^{\circ} \text{C}$ to ${0}^{\circ} \text{C}$
2nd - melt ice at ${0}^{\circ} \text{C}$
3rd - warm water from ${0}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$
4th - vaporize water to steam at ${100}^{\circ} \text{C}$
5th - heat steam from ${100}^{\circ} \text{C}$ to ${115}^{\circ} \text{C}$

1st.
$\text{heat" = "75 g" * 2.09J/("g" * ^@"C") * 5^@"C" = "0.784 kJ}$

2nd
$\text{heat" = "75 g"/(18"g"/"mol") * (6.01"kJ"/"mole") = "25.04 kJ}$

3rd
$\text{heat" = "75 g" * 4.184 "J"/("g" * ^@"C") * 100^@"C" = "31.38 kJ}$

4th
$\text{heat" = "75 g"/(18"g"/"mol") * 40.79 "kJ"/"mole" = "169.9 kJ}$

5th
$\text{heat" = "75 g" * 1.85 "J"/("g" * ^@"C") * 15^@"C" = "2.08 kJ}$

And to get the total heat, you add them all up.
About $\text{229.2 kJ}$.