Question

What is the mass of an iron sample that absorbed 1048 J of heat to raise the temperature of iron [Specific Heat - 0.448 J/(g°C)] from 100.°C to 200.°C?

Answer 1

Answer:

$23.4\text{ g}$

Explanation:

Here, we want to get the mass of the iron

Mathematically:

$Q\text{ = mc}\Delta T$

where:

m is the mass of iron that we want to calculate

c is the specific heat capacity which is 0.448 J/g°C

delta T is the change in temperature which is (200-100 = 100°C)

Q is 1048 J

Substituting the values, we have it that:

$\begin{gathered} 1048\text{ = m }\times\text{ 0.448 }\times\text{ 100} \\ m\text{ =}\frac{1048}{100\times0.448} \\ m\text{ = 23.4 g} \end{gathered}$