Animated Solution for Chemistry - Chemical Thermodynamics: For silver, $C_p({\text{J K}}^{-1}{\text{mol}}^{-1})=23+0.01T$. If the temperature ($T$) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of $\Delta H$ will be close to

Imagine you have a block of silver, and you start heating it up. You might intuitively think that adding a certain amount of heat will always raise its temperature by the exact same amount. But nature is a bit more complicated than that!

As the silver gets hotter, its atoms vibrate more intensely, and it actually becomes slightly harder to raise its temperature further. This means its heat capacity, $C_p$, isn't a constant number—it increases with temperature.

In our problem, the molar heat capacity is given by the temperature-dependent function:

We need to find the total heat added at constant pressure. In thermodynamics, the heat exchanged at constant pressure is exactly what we call the change in enthalpy, $\Delta H$.

Kirchhoff's Law

Because the heat capacity is constantly changing as the temperature rises, we cannot just use the simple formula $\Delta H=n{C}_p\Delta T$. That elementary formula only works if the heat capacity is perfectly constant!

Instead, we have to use the power of calculus to account for the continuous change. We need to add up all the infinitesimal amounts of heat required for each tiny increase in temperature.

This means we need to integrate. The total enthalpy change is the integral of the number of moles, $n$, times the molar heat capacity, $C_p$, with respect to temperature:

Let's set up our integral with the given values. We are given that we have $3$ moles of silver, so $n=3$.

We are heating the silver from an initial temperature of $300\text{ K}$ to a final temperature of $1000\text{ K}$. These will serve as the lower and upper limits of our definite integral.

Substituting our temperature-dependent function for $C_p$, we get a clear mathematical path forward:

Now for the fun part—performing the integration! We can pull the constant $3$ outside the integral to keep our calculation clean.

Inside, we integrate term by term. The integral of the constant $23$ is simply $23T$. For the second term, we use the power rule, making the integral of $0.01T$ equal to $\frac{0.01T^2}{2}$.

Putting this together, we are ready to apply our limits:

It's time to apply the upper and lower limits. A quick tip here: instead of calculating the whole expression twice, you can apply the limits to each term individually.

For the first term, we have $23(1000-300)$. For the second term, we have $0.005(1000^2-300^2)$.

Watch out for silly mistakes here—it is the difference of the squares, not the square of the difference!

Let's crunch these numbers carefully. $1000-300$ is $700$, and $23\times 700$ gives us $16100$.

Now for the squares: $1000^2$ is $1,000,000$, and $300^2$ is $90,000$. Subtracting those gives us $910,000$.

Multiplying $910,000$ by $0.005$ gives us $4550$. So, inside our brackets, we have:

We are almost at the finish line! Adding the terms inside the bracket gives us $20650$.

Multiplying this by the $3$ moles we left outside gives us exactly $61950\text{ J}$.

Looking at our options, they are all in kilojoules. Dividing by $1000$, we get $61.95\text{ kJ}$, which is incredibly close to $62\text{ kJ}$. This matches our first option perfectly!