For silver, `C_(P)(J K^(-1)"mol"^(-1))=23+0.01T`. If the temperature `(T)` of `3` moles of silver is raised from `300K` to `1000K` at `1` atm pressure, the value of `DeltaH` will be close to :

Options

A. `62kJ`
B. `16kJ`
C. `21kJ`
D. `13kJ`

Answer

To solve the problem, we need to calculate the change in enthalpy (ΔH) for 3 moles of silver when the temperature is raised from 300 K to 1000 K, given the specific heat capacity (Cp) as a function of temperature (T).

### Step-by-Step Solution:

1. **Identify the given data:**
- Specific heat capacity of silver: \( C_p = 23 + 0.01T \) (in J K\(^{-1}\) mol\(^{-1}\))
- Number of moles of silver, \( n = 3 \) moles
- Initial temperature, \( T_1 = 300 \) K
- Final temperature, \( T_2 = 1000 \) K

2. **Set up the formula for change in enthalpy (ΔH):**
The change in enthalpy can be calculated using the formula:
\[ \Delta H = n \int_{T_1}^{T_2} C_p \, dT \]

3. **Substitute the expression for Cp into the integral:**
\[ \Delta H = 3 \int_{300}^{1000} (23 + 0.01T) \, dT \]

4. **Calculate the integral:**
We can break this integral into two parts:
\[ \Delta H = 3 \left( \int_{300}^{1000} 23 \, dT + \int_{300}^{1000} 0.01T \, dT \right) \]

- For the first part:
\[ \int_{300}^{1000} 23 \, dT = 23(T) \bigg|_{300}^{1000} = 23(1000 - 300) = 23 \times 700 = 16100 \, \text{J} \]

- For the second part:
\[ \int_{300}^{1000} 0.01T \, dT = 0.01 \left( \frac{T^2}{2} \right) \bigg|_{300}^{1000} = 0.01 \left( \frac{1000^2}{2} - \frac{300^2}{2} \right) \]
\[ = 0.01 \left( \frac{1000000}{2} - \frac{90000}{2} \right) = 0.01 \left( 500000 - 45000 \right) = 0.01 \times 455000 = 4550 \, \text{J} \]

5. **Combine the results:**
Now, we can substitute back into the ΔH equation:
\[ \Delta H = 3 \left( 16100 + 4550 \right) = 3 \times 20550 = 61650 \, \text{J} \]

6. **Convert to kJ:**
\[ \Delta H = 61.65 \, \text{kJ} \approx 62 \, \text{kJ} \]

### Final Answer:
The value of ΔH will be close to **62 kJ**.