Question #779cf

1 Answer
Mar 1, 2017

Answer:

#"19,500 J"#

Explanation:

In order to be able to solve this problem, you must know the specific heat of water, which you'll find listed as

#c_"water" = "4.18 J g"^(-1)""^@"C"^(-1)#

Now, the specific heat of a given substance tells you the amount of heat needed to increase the temperature of #"1 g"# of said substance by #1^@"C"#.

In the case of water, you have

#c_"water" = color(blue)("4.18 J") color(white)(.)color(red)("g"^(-1) color(white)(.)color(purple)(""^@"C"^(-1))#

which means that you need #color(blue)("4.18 J")# of heat to raise the temperature of #color(red)("1 g")# of water by #1^@"C"#.

Now, you need to raise the temperature of #"85.0 g"# of water by

#DeltaT = 80.0^@"C" - 25.0^@"C" = 55.0^@"C"#

The first thing to do here is to calculate how much heat will be needed to raise the temperature of #"85.0 g"# of water by #1^@"C"#. To do that, use the specific heat of water as a conversion factor

#85.0 color(red)(cancel(color(black)("g"))) * "4.18 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "355.3 J" ""^@"C"^(-1)#

This means that for every #1^@"C"# increase in the temperature of an #"85.0-g"# sample of water, you need #"355.3 J"# of heat.

In your case, you will need

#55.0 color(red)(cancel(color(black)(""^@"C"))) * "355.3 J"/(1color(red)(cancel(color(black)(""^@"C")))) = color(darkgreen)(ul(color(black)("19,500 J")))#

The answer is rounded to three sig figs.

You can double-check this result by using the equation

#color(blue)(ul(color(black)(q = m * c * DeltaT)))#

Here

  • #q# is the heat lost or gained by the substance
  • #m# is the mass of the sample
  • #c# is the specific heat of the substance
  • #DeltaT# is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

In your case, you will have

#q = 85.0 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 55.0 color(red)(cancel(color(black)(""^@"C")))#

#q = color(darkgreen)(ul(color(black)("19,500 J")))#