The motor operates at 100 V and draws 1 A, so the total electrical power input is:

$$P_{\text{in}} = IV = 100 \, \text{V} \times 1 \, \text{A} = 100 \, \text{W}.$$

With an efficiency of 91.6%, only 91.6% of the input power is used for the motor's work, while the rest is lost as heat. The output power is:

$$P_{\text{out}} = 0.916 \times 100 \, \text{W} = 91.6 \, \text{W}.$$

The power lost is:

$$P_{\text{loss}} = P_{\text{in}} - P_{\text{out}} = 100 \, \text{W} - 91.6 \, \text{W} = 8.4 \, \text{W}.$$

Since $1 \, \text{W} = 1 \, \text{J/s}$ and $1 \, \text{cal} \approx 4.2 \, \text{J}$, we convert the lost power into calories per second:

$$P_{\text{loss}} (\text{in cal/s}) = \frac{8.4 \, \text{J/s}}{4.2 \, \text{J/cal}} = 2 \, \text{cal/s}.$$

Thus, the correct answer is:

Option B: 2