Question #2c289

1 Answer
Mar 3, 2017

Answer:

#"23,000 J"#

Explanation:

Your tool of choice here will be 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, the change in temperature is equal to #66^@"C"#. In other words, the difference between the final and the initial temperatures of the water is equal to #66^@"C"#.

Now, look up the specific heat for water. You should find it listed as

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

The specific heat of a substance is a measure of how much energy is needed to increase the temperature of #"1 g"# of said substance by #1^@"C"#.

So, plug in your values to find

#q = 83 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 66color(red)(cancel(color(black)(""^@"C")))#

#q = color(darkgreen)(ul(color(black)("23,000 J")))#

The answer must be rounded to two sig figs, the number of significant figures you have for the mass of the sample and for the change in temperature.