# Question 95b88

Jun 9, 2017

$\text{75,000 J}$

#### Explanation:

The trick here is to realize that you must keep track of two phase changes as you go from ice at ${0}^{\circ} \text{C}$ to steam at ${100}^{\circ} \text{C}$.

More specifically, you have

• ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$
• liquid water at ${0}^{\circ} \text{C}$ to liquid water at ${100}^{\circ} \text{C}$
• liquid water at ${100}^{\circ} \text{C}$ to steam at ${100}^{\circ} \text{C}$

In order to be able to calculate the heat required to convert your sample, you need to know

• DeltaH_"fus" = color(blue)("333.55 J g"^(-1)) -> the enthalpy of fusion of water
• c_"water" = color(purple)("4.18 J g"^(-1)""^@"C"^(-1)) -> the specific heat of water
• DeltaH_"vap" = color(darkorange)("2257 J g"^(-1)) -> the enthalpy of vaporization of water

So, the first thing to calculate here is the heat required to go from ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$. To do that, use the enthalpy of fusion of water, which tells you the enthalpy change that accompanies the conversion of $\text{1 g}$ of ice to liquid water at its normal melting point.

25 color(red)(cancel(color(black)("g ice"))) * color(blue)("333.55 J"/(1color(red)(cancel(color(blue)("g ice"))))) = "8,338.75 J"

Next, calculate the heat needed to heat your sample from liquid water at ${0}^{\circ} \text{C}$ to liquid water at ${100}^{\circ} \text{C}$.

The speciifc heat of water tells you that you need $\textcolor{p u r p \le}{\text{4.18 J}}$ of heat to increase the temperature of $\textcolor{p u r p \le}{\text{1 g}}$ of liquid water by $\textcolor{p u r p \le}{{1}^{\circ} \text{C}}$.

This means that in order to increase the temperature of $\text{25 g}$ of water, you need

25 color(red)(cancel(color(black)("g"))) * color(purple)("4.18 J")/(color(purple)(1)color(red)(cancel(color(purple)("g"))) * color(purple)(1^@"C")) = "104.5 J"^@"C"^(-1)

So in order to increase the temperature of $\text{25 g}$ of water by ${1}^{\circ} \text{C}$, you need to provide it with $\text{104.5 J}$. In your case, the temperature change is

${100}^{\circ} \text{C" - 0^@"C" = 100^@"C}$

so you will need a total of

100color(red)(cancel(color(black)(""^@"C"))) * "104.5 J"/(1color(red)(cancel(color(black)(""^@"C")))) = "10,450 J"

Finally, to convert your sample from liquid water at ${100}^{\circ} \text{C}$ to vapor at ${100}^{\circ} \text{C}$, use the enthalpy of vaporization of water.

In this case, you know that you need $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{2257 J}}$ of heat to convert $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{1 g}}$ of water from liquid to vapor at its normal boling point, which means that your sampel will need

25 color(red)(cancel(color(black)("g"))) * color(darkorange)("2257 J"/(1color(red)(cancel(color(darkorange)("g"))))) = "56,425 J"#

Finally, add all the three heats to get the total heat needed

$\text{total heat" = "8,338.75 J" + "10,450 J" + "56,425 J}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{total heat = 75,000 J}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the two temperatures.