C

To calculate the quantity of heat required to raise the temperature of 216 g of silver from 50 K to 100 K, we will first need to calculate the moles of silver and then integrate the specific heat equation over the desired temperature range.
Step 1: Calculate the number of moles of silver (Ag). The molar mass of silver is approximately 107.87 g/mol.
\[ \text{moles of Ag} = \frac{216 \text{ g}}{107.87 \text{ g/mol}} \approx 2.00 \text{ mol} \]
Step 2: Define the specific heat capacity formula:
\[ c_p(T) = 0.076 T - 0.00026 T^2 - 0.15 \]
Step 3: The heat (q) required to raise the temperature is given by the integral:
\[ q = n \int_{T_1}^{T_2} c_p(T) \, dT \]
In our case, \( n = 2.00 \) mol, \( T_1 = 50 \) K, and \( T_2 = 100 \) K:
\[ q = 2.00 \int_{50}^{100} (0.076 T - 0.00026 T^2 - 0.15) \, dT \]
Step 4: Compute the integrated function:
\[ \int (0.076 T - 0.00026 T^2 - 0.15) \, dT = 0.038 T^2 - \frac{0.00026}{3} T^3 - 0.15 T + C \]
Step 5: Calculate the definite integral from 50 to 100:
\[ q = 2.00 \left[ \left(0.038(100^2) - \frac{0.00026}{3}(100^3) - 0.15(100) \right) - \left(0.038(50^2) - \frac{0.00026}{3}(50^3) - 0.15(50) \right) \right] \]
Step 6: Evaluate substituting values:
First evaluation at T = 100:
\[ 0.038(10000) - \frac{0.00026}{3}(1000000) - 15 = 380 - 86.67 - 15 = 278.33 \]
For T = 50:
\[ 0.038(2500) - \frac{0.00026}{3}(125000) - 7.5 = 95 - 1.083 - 7.5 = 86.417 \]
Finally calculating:
\[ q = 2.00(278.33 - 86.417) = 2.00 imes 191.913 = 383.826 \text{ cal} \]
Therefore, the quantity of heat required is approximately 383.83 cal, which indicates option C.