# 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?

Aug 14, 2018

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:

$Q = c \cdot m \cdot \Delta T$

The energy required to raise the temperature of $m = 45.6 \textcolor{w h i t e}{l} \text{g}$ of lead by $\Delta T = 13.3 \textcolor{w h i t e}{l} \text{^"o""C}$ is $78.2 \textcolor{w h i t e}{l} \text{J}$ according to the question;

• $Q = 78.2 \textcolor{w h i t e}{l} \text{J}$,
• $m = 45.6 \textcolor{w h i t e}{l} \text{g}$,
• $\Delta T = 13.3 \textcolor{w h i t e}{l} \text{^"o""C}$,
• 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")

$c = \left(78.2 \textcolor{w h i t e}{l} \text{J") // (45.6 color(white)(l) "g" * 13.3 color(white)(l) ""^"o""C}\right)$
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

$Q = c \cdot m \cdot \Delta T$ $\implies$ $c = \frac{Q}{m \cdot \Delta T}$