# Question #5141e

##### 1 Answer

#### Explanation:

The idea here is that you're going to need enough heat to

- Heat
#"4.80 g"# of ice at#-30.0^@"C"# to ice at#0^@"C"# - Convert the sample from ice at
#0^@"C"# to liquid water at#0^@"C"# , i.e. perform a solid#-># liquidphase change- Heat
#"4.80 g"# of liquid water at#0^@"C"# to liquid water at#100^@"C"# - Convert the sample from liquid water at
#100^@"C"# to vapor at#100^@"C"# , i.e. perform a liquid#-># vaporphase change

The **total heat** needed to get the sample from ice at **sum** of the heat needed to ensure that every step in this series takes place.

#q_"total" = sumq_"each step"#

Now, for each step, you will need to use

- The specific heat of ice
- The heat of fusion of water
- The specific heat of water
- The heat of vaporization of water

The two equations that you will use are

#color(blue)(ul(color(black)(q = m * DeltaH_"fus/vap")))" " " "color(darkorange)((1))# This equation is used for

phase changes, i.e. when the change in temperature is equal to zero. Here#m# is themassof the sample and#DeltaH_"fus/vap"# is the heat of fusion.vaporization, depending on which phase change you're dealing with.

#color(blue)(ul(color(black)(q = m * c_"ice/water" * DeltaT)))" " " "color(darkorange)((2))# This equation is used when you're

heatingthe samplewithoutperforming a phase change. Here#c# is thespecific heatof ice/water and#DeltaT# is thechange in temperature

This means that, for each step, you will use

- Use equation
#color(darkorange)((2))# with#c_"ice"# and a change in temperature of#DeltaT = 0^@"C" - (-30.0^@"C") = 30.0^@"C"# - Use equation
#color(darkorange)((1))# with#DeltaH_"fus"# - Use equation
#color(darkorange)((2))# with#c_"water"# and a change in temperature of#DeltaT = 100^@"C" - 0^@"C" = 100^@"C"# - Use equation
#color(darkorange)((1))# with#DeltaH_"vap"#

So, you can put all this together to get

#q_"total" = overbrace(4.80 color(red)(cancel(color(black)("g"))) * "2.06 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 30.0color(red)(cancel(color(black)(""^@"C"))))^(color(blue)("step 1")) + overbrace(4.80 color(red)(cancel(color(black)("g"))) * "334.16 J" color(red)(cancel(color(black)("g"^(-1)))))^(color(blue)("step 2"))#

#color(white)(q_"total") + overbrace(4.80 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 100.0color(red)(cancel(color(black)(""^@"C"))))^(color(blue)("step 3")) + overbrace(4.80 color(red)(cancel(color(black)("g"))) * "2259J" color(red)(cancel(color(black)("g"^(-1)))))^(color(blue)("step 4"))#

#color(darkgreen)(ul(color(black)(q_"total" = "14,800 J")))#

The answer is rounded to three **sig figs**.