Question #95b88

1 Answer
Jun 9, 2017

Answer:

#"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^@"C"# to steam at #100^@"C"#.

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)) -># 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^@"C"# to liquid water at #0^@"C"#. To do that, use the enthalpy of fusion of water, which tells you the enthalpy change that accompanies the conversion of #"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^@"C"# to liquid water at #100^@"C"#.

The speciifc heat of water tells you that you need #color(purple)("4.18 J")# of heat to increase the temperature of #color(purple)("1 g")# of liquid water by #color(purple)(1^@"C")#.

This means that in order to increase the temperature of #"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 #"25 g"# of water by #1^@"C"#, you need to provide it with #"104.5 J"#. In your case, the temperature change is

#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 #100^@"C"# to vapor at #100^@"C"#, use the enthalpy of vaporization of water.

In this case, you know that you need #color(darkorange)("2257 J")# of heat to convert #color(darkorange)("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

#"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.