# What is the specific heat capacity of a 50-gram piece of 100°C metal that will change 400 grams of 20°C water to 22°C?

May 20, 2016

${s}_{{\text{metal")=0.86J/(g*}}^{\circ} C}$

#### Explanation:

According to the law of conservation of energy that sates:
Energy cannot be created nor destroyed, we can conclude that the energy lost from the metal is absorbed by water, and therefore;

${q}_{l o s t} = {q}_{\text{gained}}$

in here, the metal will lose heat and water will gain heat. Note that the specific heat capacity of water is s=4.18J/(g*""^@C).

Thus, $m \times {s}_{\text{metal}} \times \Delta T = m \times {s}_{w a t e r} \times \Delta T$

50cancel(g)xxs_("metal")xx(100cancel(""^@C)-22cancel(""^@C))=400cancel(g)xx4.18J/(g*""^@C)xx(22cancel(""^@C)-20cancel(""^@C))

Now we can solve for the specific heat capacity of the metal and we get:

$\implies {s}_{{\text{metal")=0.86J/(g*}}^{\circ} C}$

Here is a video that explains this topic with more details:
Thermochemistry | Enthalpy and Calorimetry.