# Question 34abf

Jul 27, 2017

$\text{7000 cal}$

#### Explanation:

The idea here is that you need to take into account four distinct heats

• the heat required to turn $\text{10 g}$ of ice at its normal melting point of ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$
• the heat required to heat $\text{10 g}$ of liquid water from ${0}^{\circ} \text{C}$to its normal boiling point of ${100}^{\circ} \text{C}$
• the heat required to convert $\text{10 g}$ of liquid water at ${100}^{\circ} \text{C}$ to vapor at ${100}^{\circ} \text{C}$
• the heat required to heat $\text{10 g}$ of steam from ${100}^{\circ} \text{C}$ to ${110}^{\circ} \text{C}$

In your case, you will have

10 color(red)(cancel(color(black)("g ice"))) * overbrace("80 cal"/(1color(red)(cancel(color(black)("g ice")))))^(color(blue)("the latent heat of fusion of water")) = "800 cal"

This represents the heat needed to convert your sample from ice at ${0}^{\circ} \text{C}$ to liquid water at ${100}^{\circ} \text{C}$, or ${q}_{1}$.

Now, in order to figure out how much heat is needed to heat the liquid water from ${0}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$, you need to know the specific heat of water, which you can take as

${c}_{\text{water" = "1 cal g"^(-1)""^@"C}}^{- 1}$

In your case, you will need

10 color(red)(cancel(color(black)("g"))) * overbrace("1 cal"/(1color(red)(cancel(color(black)("g"))) * 1^@"C"))^(color(blue)("the specific heat of water")) = "10 cal"""^@"C"^(-1)

to increase the temperature of the sample by ${1}^{\circ} \text{C}$, which implies that in order to increase its temperature by ${100}^{\circ} \text{C}$, you will need

100 color(red)(cancel(color(black)(""^@"C"))) * overbrace("10 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 10 g of water")) = "1000 cal"

This represents the total heat need to increase the temperature of $\text{10 g}$ of liquid water from ${0}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$, or ${q}_{2}$.

Next, you will have

10 color(red)(cancel(color(black)("g"))) * overbrace("540 cal"/(1 color(red)(cancel(color(black)("g")))))^(color(blue)("the latent heat of vaporization of water")) = "5400 cal"

This represents the heat needed t convert $\text{10 g}$ of liquid water from ${100}^{\circ} \text{C}$ to vapor at ${100}^{\circ} \text{C}$, or ${q}_{3}$.

Finally, you can increase the temperature of $\text{10 g}$ of steam by adding

10 color(red)(cancel(color(black)("g steam"))) * overbrace("0.5 cal"/(1color(red)(cancel(color(black)("g steam"))) * 1^@"C"))^(color(blue)("the specific heat of steam")) = "5 cal"""^@"C"^(-1)

which implies that in order to increase the temperature of $\text{10 g}$ of steam by

${110}^{\circ} \text{C" - 100^@"C" = 10^@"C}$

you will need

10 color(red)(cancel(color(black)(""^@"C"))) * overbrace("5 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 10 g of steam")) = "50 cal"

This represents the heat needed to increase the temperature of $\text{10 g}$ of steam from ${100}^{\circ} \text{C}$ to ${110}^{\circ} \text{C}$, or ${q}_{4}$.

You can thus say that you have

${q}_{\text{total}} = {q}_{1} + {q}_{2} + {q}_{3} + {q}_{4}$

which, in your case, is equal to

${q}_{\text{total" = "800 cal + 1000 cal + 5400 cal + 50 cal}}$

q_"total" = color(darkgreen)(ul(color(black)("7000 cal")))#

The answer must be rounded to one significant figure, the number of sig figs you have for the mass of ice.