# Question #95b88

##### 1 Answer

#### Explanation:

The trick here is to realize that you must keep track of two *phase changes* as you go from ice at

More specifically, you have

ice at#0^@"C"# to liquid water at#0^@"C"# liquid water at#0^@"C"# to liquid water at#100^@"C"# liquid water at#100^@"C"# to steam at#100^@"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)) -># theenthalpy of fusionof water#c_"water" = color(purple)("4.18 J g"^(-1)""^@"C"^(-1)) -># thespecific heatof water#DeltaH_"vap" = color(darkorange)("2257 J g"^(-1)) -># theenthalpy of vaporizationof water

So, the first thing to calculate here is the heat required to go from ice at

#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

The speciifc heat of water tells you that you need

This means that in order to increase the temperature of

#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

#100^@"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

In this case, you know that you 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

#"total heat" = "8,338.75 J" + "10,450 J" + "56,425 J"#

#color(darkgreen)(ul(color(black)("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.