Fill in the blanks:
$(a)$ $0.49 \frac{\text{cal}}{\text{cm} \cdot \text{K} \cdot \text{s}} = \dots \frac{\text{J}}{\text{m} \cdot \text{K} \cdot \text{s}}$
$(b)$ If the rate of emission of heat of a substance is less than its rate of absorption,then its temperature $\dots$.
$(c)$ The rate of emission of heat of a substance is directly proportional to $\dots$ of temperature of it and surroundings.

(N/A) Given $1 \text{ cal} = 4.184 \text{ J}$ and $1 \text{ cm} = 10^{-2} \text{ m}$.
$0.49 \frac{\text{cal}}{\text{cm} \cdot \text{K} \cdot \text{s}} = 0.49 \times \frac{4.184 \text{ J}}{10^{-2} \text{ m} \cdot \text{K} \cdot \text{s}} = 0.49 \times 418.4 \approx 205 \frac{\text{J}}{\text{m} \cdot \text{K} \cdot \text{s}}$.
Note: If using $1 \text{ cal} = 4.2 \text{ J}$,then $0.49 \times 420 = 205.8 \approx 206$.
$(b)$ If the rate of absorption is greater than the rate of emission,the substance gains net energy,so its temperature will increase.
$(c)$ According to Newton's Law of Cooling,the rate of loss of heat is directly proportional to the difference in temperature between the body and its surroundings,provided the difference is small.