Step 1 - How specific heat relates to heat
The heat absorbed by a substance can be related to its mass (m) and temperature (delta T) variation by the following formula:
[tex]Q=mc\Delta T[/tex]
We will use the following formula to solve the poblem.
Step 2 - Solving the exercise
According to this exercise:
[tex]\begin{gathered} Q=2500J \\ \Delta T=350-300=50°C \\ m=347.2g \\ c=? \end{gathered}[/tex]
Setting the values in the equation:
[tex]\begin{gathered} 2500=300\times347.2\times c \\ \\ c=\frac{2500}{104160}=0.024J/g°C \end{gathered}[/tex]
Answer: the specific heat of this metal is 0.024 J/g.°C
