Question #34abf

1 Answer
Jul 27, 2017

Answer:

#"7000 cal"#

Explanation:

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

  • the heat required to turn #"10 g"# of ice at its normal melting point of #0^@"C"# to liquid water at #0^@"C"#
  • the heat required to heat #"10 g"# of liquid water from #0^@"C"#to its normal boiling point of #100^@"C"#
  • the heat required to convert #"10 g"# of liquid water at #100^@"C"# to vapor at #100^@"C"#
  • the heat required to heat #"10 g"# of steam from #100^@"C"# to #110^@"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^@"C"# to liquid water at #100^@"C"#, or #q_1#.

Now, in order to figure out how much heat is needed to heat the liquid water from #0^@"C"# to #100^@"C"#, you need to know the specific heat of water, which you can take as

#c_"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^@"C"#, which implies that in order to increase its temperature by #100^@"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 #"10 g"# of liquid water from #0^@"C"# to #100^@"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 #"10 g"# of liquid water from #100^@"C"# to vapor at #100^@"C"#, or #q_3#.

Finally, you can increase the temperature of #"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 #"10 g"# of steam by

#110^@"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 #"10 g"# of steam from #100^@"C"# to #110^@"C"#, or #q_4#.

You can thus say that you have

#q_"total" = q_1 + q_2 + q_3 + q_4#

which, in your case, is equal to

#q_"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.