For an element, `Cp = 23 + 0.01 T("JK"^(-1)"mol"^(-1))` . If temperature of 3 moles of that element is raised from 300 K to 1000 K at 1 atm pressure, the value of `DeltaH` will be ______ kJ.

Explanation

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

### Step-by-Step Solution:

1. **Identify the given values**:
- Cp = 23 + 0.01T (in J/(K·mol))
- Number of moles (n) = 3 moles
- Initial temperature (T1) = 300 K
- Final temperature (T2) = 1000 K

2. **Write the formula for ΔH**:
\[ \Delta H = n \int_{T_1}^{T_2} C_p \, dT \]
Here, we can take the number of moles (n) out of the integral.

3. **Substitute Cp into the formula**:
\[ \Delta H = n \int_{T_1}^{T_2} (23 + 0.01T) \, dT \]
\[ \Delta H = 3 \int_{300}^{1000} (23 + 0.01T) \, dT \]

4. **Separate the integral**:
\[ \Delta H = 3 \left( \int_{300}^{1000} 23 \, dT + \int_{300}^{1000} 0.01T \, dT \right) \]

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

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

7. **Combine the results**:
\[ \Delta H = 3 \left( 16100 + 4550 \right) = 3 \times 20550 = 61650 \, \text{J} \]

8. **Convert J to kJ**:
\[ \Delta H = \frac{61650}{1000} = 61.65 \, \text{kJ} \]

### Final Answer:
\[ \Delta H = 61.65 \, \text{kJ} \]