# 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

#### Answer:

#### Explanation:

By definition, the specific heat **energy** required to raise the temperature of **one gram** of the substance by **one degree Celsius**. The energy

#Q = c * m * Delta T#

The energy required to raise the temperature of

#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

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