For silver, $$C_p$$ (J K$$^{-1}$$ 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:

We have to calculate the enthalpy change at constant pressure. The basic thermodynamic relation is stated first:

$$\Delta H = n\int_{T_1}^{T_2} C_p\,dT$$

Here $$n = 3\;\text{mol},\; T_1 = 300\;\text{K},\; T_2 = 1000\;\text{K},$$ and the temperature-dependent molar heat capacity of silver is given as

$$C_p = 23 + 0.01\,T \quad \bigl(\text{J K}^{-1}\text{ mol}^{-1}\bigr).$$

Now we substitute this expression for $$C_p$$ inside the integral:

$$\Delta H = 3\int_{300}^{1000} \bigl(23 + 0.01\,T\bigr)\,dT.$$

We split the integral term by term:

$$\Delta H = 3\left[\int_{300}^{1000} 23\,dT + \int_{300}^{1000} 0.01\,T\,dT\right].$$

Next we integrate each part algebraically. For the first integral, $$\int 23\,dT = 23\,T.$$ For the second, we use the power-rule $$\int T\,dT = \dfrac{T^2}{2},$$ so $$\int 0.01\,T\,dT = 0.01\,\dfrac{T^2}{2}.$$

Writing the definite integrals explicitly:

$$\Delta H = 3\left[23\,T\Bigl|_{300}^{1000} + 0.01\,\dfrac{T^2}{2}\Bigl|_{300}^{1000}\right].$$

We evaluate the first bracket:

$$23\,T\Bigl|_{300}^{1000} = 23(1000) - 23(300) = 23(1000-300) = 23 \times 700 = 16100\;\text{J mol}^{-1}.$$

For the second bracket we first compute the squares:

$$T^2\Bigl|_{300}^{1000} = 1000^2 - 300^2 = 1000000 - 90000 = 910000.$$

Then

$$0.01\,\dfrac{T^2}{2}\Bigl|_{300}^{1000} = 0.01 \times \dfrac{910000}{2} = 0.01 \times 455000 = 4550\;\text{J mol}^{-1}.$$

Adding the two contributions for one mole:

$$16100\;\text{J} + 4550\;\text{J} = 20650\;\text{J mol}^{-1} = 20.65\;\text{kJ mol}^{-1}.$$

Finally, we multiply by the number of moles:

$$\Delta H = 3 \times 20.65\;\text{kJ} = 61.95\;\text{kJ} \approx 62\;\text{kJ}.$$

This numerical value lies closest to the option 62 kJ.

Hence, the correct answer is Option B.