# Question f8771

Jul 14, 2015

You'd need +6.79 kJ.

#### Explanation:

In order to determine how much heat is needed to melt that much cobalt at its melting point you need to know two things

• the mass of the sample;
• the heat of fusion, $\Delta {H}_{\text{fus}}$, of cobalt.

A substance's enthalpy of fusion (heat of fusion) tells you how much energy is needed to convert one gram of that substance from solid to liquid, or given off when one gram is converted from liquid to solid.

Since you're essentially dealing with a phase change, the temperature of the sample will remain unchanged.

Notice that the value of $\Delta {H}_{\text{fus}}$ given to you is expressed in kJ per mole. This means that you're actually dealing with a molar enthalpy of fusion.

In other words, 15.5 kJ of heat are needed to melt 1 mole, not one gram, of cobalt at its melting point of $1495 \text{^@"C}$.

The equation you'll use looks like this

$q = n \cdot \Delta {H}_{\text{fus}}$, where

$n$ - the number of moles of cobalt.

To calculte $n$, use cobalt's molar mass

25.8cancel("g") * "1 mole Co"/(58.933cancel("g")) = "0.4378 moles Co"

This means that you'll need

$q = n \cdot \Delta {H}_{\text{fus}}$

q = 0.4378cancel("moles") * 15.5"kJ"/cancel("mole") = "+6.7859 kJ"#

of heat to melt your sample.

Rounded to three sig figs, the answer will be

$q = \textcolor{g r e e n}{\text{+6.79 kJ}}$