The specific heat capacity at constant pressure (\(C_p\)) of a solid metal is given by \[ C_p\ (\text{in J/mol-K}) = 20 + (5\times10^{-3})T \] valid for \(T = 298\) K to \(1000\) K. At constant pressure, if the temperature of \(2\) moles of the metal is increased from \(300\) K to \(600\) K, find the change in enthalpy of the metal (answer as an integer), in Joules.
Show Hint
Integrate \(C_p\) with respect to \(T\) and multiply by the number of moles: \(\Delta H=n\int C_p\,dT\).
Step 1: Split $C_p$ into a constant part and a temperature dependent part.
\[ C_p=20+(5\times10^{-3})T \]
The enthalpy change is $\Delta H=n\displaystyle\int_{300}^{600}C_p\,dT$, and since the integral is linear we can integrate each part on its own and add the results.
Step 2: Integrate the constant term.
\[ \int_{300}^{600}20\,dT=20\times(600-300)=20\times300=6000\ \text{J/mol} \]
Step 3: Integrate the linear term.
\[ \int_{300}^{600}5\times10^{-3}T\,dT=\frac{5\times10^{-3}}{2}\Big[T^2\Big]_{300}^{600}=2.5\times10^{-3}\left(600^2-300^2\right) \]
Now $600^2=360000$ and $300^2=90000$, so the difference is $270000$.
\[ 2.5\times10^{-3}\times270000=675\ \text{J/mol} \]
Step 4: Add the two pieces to get the molar enthalpy change.
\[ \int_{300}^{600}C_p\,dT=6000+675=6675\ \text{J/mol} \]
Step 5: Scale by the number of moles.
\[ \Delta H=n\times6675=2\times6675=13350\ \text{J} \]
This agrees with the direct integration method and lands inside the accepted range of $13340$ to $13360$ J.
\[ \boxed{\Delta H=13350\ \text{J}} \]
