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Step-by-Step Solution
Step 1: Identify Given Data
• Number of moles, $n = 3$
• Initial temperature, $T_{1} = 300\,\mathrm{K}$
• Final temperature, $T_{2} = 1000\,\mathrm{K}$
• Heat capacity (at constant pressure) for silver, $C_{p} = 23 + 0.01T\,(\mathrm{J\,K^{-1}\,mol^{-1}})$
Step 2: Recall the Expression for Enthalpy Change
To find the enthalpy change $\Delta H$ for heating from $T_{1}$ to $T_{2}$ at constant pressure, we use:
$$
\Delta H = \int_{T_1}^{T_2} n \, C_{p}\,dT.
$$
Step 3: Substitute $C_{p}$ into the Integral
Given $C_{p} = 23 + 0.01T$, substitute into the integral:
$$
\Delta H = \int_{300}^{1000} 3 \left( 23 + 0.01T \right) \, dT.
$$
Step 4: Simplify the Integral
Factor out the constant 3:
$$
\Delta H = 3 \int_{300}^{1000} \left( 23 + 0.01T \right) \, dT.
$$
Inside the integral:
$$
\int \left( 23 + 0.01T \right) \, dT
= 23T + 0.01 \times \frac{T^2}{2}
= 23T + 0.005\, T^2.
$$
Step 5: Evaluate the Integral From 300 K to 1000 K
Let
$$
F(T) = 23T + \frac{T^2}{200}.
$$
Then
$$
\int_{300}^{1000} \left( 23 + 0.01T \right) \, dT
= F(1000) - F(300).
$$
Calculate $F(1000)$:
$$
F(1000) = 23 \times 1000 + \frac{1000^2}{200}
= 23000 + 5000 = 28000.
$$
Calculate $F(300)$:
$$
F(300) = 23 \times 300 + \frac{300^2}{200}
= 6900 + 450 = 7350.
$$
Difference:
$$
F(1000) - F(300) = 28000 - 7350 = 20650.
$$
Hence,
$$
\int_{300}^{1000} \left( 23 + 0.01T \right) \, dT = 20650.
$$
Step 6: Multiply by the Factor Outside the Integral
Since there is a factor of 3 moles, we get:
$$
\Delta H = 3 \times 20650\,\mathrm{J} = 61950\,\mathrm{J}.
$$
Step 7: Convert to Kilojoules
$$
61950\,\mathrm{J} = 61.95\,\mathrm{kJ} \approx 62\,\mathrm{kJ}.
$$
Therefore, the enthalpy change for raising the temperature of 3 moles of silver from 300 K to 1000 K is
$$
\Delta H \approx 62\,\mathrm{kJ}.
$$
Final Answer
The value of $\Delta H$ is approximately 62 kJ.
