Question

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?

Answer 1

`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)`