Question

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.

Hint:

Integrate \(C_p\) with respect to \(T\) and multiply by the number of moles: \(\Delta H=n\int C_p\,dT\).

Answer 1

Correct Answer: 13340

Step 1: Recall how enthalpy change relates to heat capacity.
At constant pressure, the enthalpy change of a substance as its temperature rises from \(T_1\) to \(T_2\) is found by integrating the heat capacity:
\[ \Delta H = n\int_{T_1}^{T_2} C_p\, dT \]
where \(n\) is the number of moles and \(C_p\) is the molar heat capacity at constant pressure.

Step 2: Write down the given heat capacity expression and limits.
\[ C_p = 20 + (5\times10^{-3})T \quad \text{J/mol-K} \]
\[ n = 2\ \text{mol}, \quad T_1 = 300\ \text{K}, \quad T_2 = 600\ \text{K} \]

Step 3: Integrate \(C_p\) with respect to \(T\).
\[ \int C_p\, dT = \int \left(20 + 5\times10^{-3}T\right) dT = 20T + \frac{5\times10^{-3}}{2}T^2 = 20T + 2.5\times10^{-3}T^2 \]

Step 4: Evaluate this expression at the upper limit \(T_2 = 600\) K.
\[ 20(600) + 2.5\times10^{-3}(600)^2 = 12000 + 2.5\times10^{-3}(360000) \]
\[ = 12000 + 900 = 12900\ \text{J/mol} \]

Step 5: Evaluate the same expression at the lower limit \(T_1 = 300\) K.
\[ 20(300) + 2.5\times10^{-3}(300)^2 = 6000 + 2.5\times10^{-3}(90000) \]
\[ = 6000 + 225 = 6225\ \text{J/mol} \]

Step 6: Subtract to get the molar enthalpy change, then scale by the number of moles.
\[ \int_{300}^{600} C_p\, dT = 12900 - 6225 = 6675\ \text{J/mol} \]
\[ \Delta H = n \times 6675 = 2 \times 6675 = 13350\ \text{J} \]

Final Answer:
The change in enthalpy of the 2 moles of metal as it heats from 300 K to 600 K is 13350 J.
\[ \boxed{\Delta H = 13350\ \text{J}} \]