Question #8e7c8

1 Answer
Jun 4, 2016

Answer:

#"39,800 J"#

Explanation:

The idea here is that you need to calculate how much heat is required in order to

  • convert your sample of ice at #-35^@"C"# to ice at #0^@"C"#
  • convert the ice at #0^@"C"# to liquid water at #0^@"C"#

In order to be able to calculate the heat required in both cases, you need to know the specific heat of ice and the heat of fusion of water

#c_"ice" = "2.06 J g"^(-1)""^@"C"^(-1)#

#DeltaH_"fus" = "334.16 J g"^(-1)#

So, a substance's specific heat tells you how much heat is needed in order to increase the temperature of #"1 g"# of that substance by #1^@"C"#.

In this case, ice has a specific heat of #"2.06 J g"^(-1)""^@"C"^(-1)#, which means that in order to increase the temperature of #"1 g"# of ice by #1^@"C"#, you need to provide it with #"2.06 J"# of heat.

Your tool of choice here will be this equation

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#, where

#q# - the amount of heat gained
#m# - the mass of the sample
#c# - the specific heat of the substance
#DeltaT# - the change in temperature, defined as the difference between the final temperature and the initial temperature

In your case, you have a change in temperature of

#DeltaT = 0^@"C" - (-35^@"C") = 35^@"C"#

Plug in your values to find

#q_1 = 98.0 color(red)(cancel(color(black)("g"))) * "2.06 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 35color(red)(cancel(color(black)(""^@"C")))#

#q_1 = "7,065.8 J"#

Now, adding this much heat to #"98.0 g"# of ice at #-35^@"C"# will get you ice at #0^@"C"#. In order to melt the ice, you must provide it with enough heat to allow it to undergo a solid #-># liquid phase change.

The heat of fusions, #DeltaH_"fus"#, tells you how much heat must be added per gram of solid ice at #0^@"C"# in order to convert it to liquid water at #0^@"C"#.

Since you have #"98.0 g"# of ice at #0^@"C"#, you will need

#98.0 color(red)(cancel(color(black)("g"))) * overbrace("334.16 J"/(1color(red)(cancel(color(black)("g")))))^(color(blue)(=DeltaH_"fus")) = "32,748 J"#

The total amount of heat needed to get your sample from ice at #-35^@"C"# to liquid water at #0^@"C"# will thus be

#q_"total" = "7,065.8 J" + "32,748 J"#

#q_"total" = "39,814 J"#

I'll leave the answer rounded to three sig figs

#"the amount of heat needed" = color(green)(|bar(ul(color(white)(a/a)color(black)("39,800 J")color(white)(a/a)|)))#