It takes 78.2 J to raise the temperature of 45.6 grams of lead by 13.3 C. What is the specific heat of lead?

1 Answer
Aug 14, 2018

Answer:

#c = 0.129 color(white)(l) "J" // ("g" * ""^"o""C")#

Explanation:

By definition, the specific heat #c# of a particular substance is the energy required to raise the temperature of one gram of the substance by one degree Celsius. The energy #Q# required to raise the temperature of a sample of that substance of mass #m# by #Delta T# would thus be:

#Q = c * m * Delta T#

The energy required to raise the temperature of #m = 45.6 color(white)(l) "g"# of lead by #Delta T = 13.3 color(white)(l) ""^"o""C"# is #78.2 color(white)(l) "J"# according to the question;

  • #Q = 78.2 color(white)(l) "J"#,
  • #m = 45.6 color(white)(l) "g"#,
  • #Delta T = 13.3 color(white)(l) ""^"o""C"#,
  • and #c# is the unknown.

In other words,

#78.2 color(white)(l) "J" = c xx (45.6 color(white)(l) "g" * 13.3 color(white)(l) ""^"o""C")#

#c = (78.2 color(white)(l) "J") // (45.6 color(white)(l) "g" * 13.3 color(white)(l) ""^"o""C")#
#color(white)(c) ~~ 0.129 color(white)(l) "J" // ("g" * ""^"o""C")#

In general, the specific heat of a substance #c# given that it takes energy #Q# to raise the temperature of a sample of mass #m# by temperature #Delta T# would be

#Q = c * m * Delta T# #=># #c = Q /( m * Delta T)#