# How much heat is required to raise the temperature from 26 degree centigrade to 50 degree centigrade for 100 gr lead?

Apr 4, 2018

$\text{400 J}$, or $\text{90 cal}$.

#### Explanation:

To solve this problem, we'll need to use this equation:

$q = m C \Delta T$

Where $q$ is heat, $m$ is mass in grams, $C$ is specific heat, and $\Delta T$ is change in temperature.

The question gives us $m$, which is $\text{100 g}$.
We can also use the initial and final temperatures given in the question, $\text{26°C}$ and $\text{50°C}$, to find $\Delta T$. $\Delta T$ is $50 - 26 = \text{24°C}$.

The specific heat of lead, from a quick search, is $\text{0.160 J/g°C}$.

So, we can just plug these into the equation and solve for $q$:

$q = m C \Delta T$
$q = \text{100 g" xx "0.160 J/g°C" xx "24°C}$
$q = \text{384 J}$ or $\text{91.78 cal}$

Since the lowest number of significant figures in the question is $1$, our answer needs to also have $1$ significant figure.
That would make it $\text{400 J}$, or $\text{90 cal}$.