Question

The emissive power of a sphere of area $0.04\text{ m}^2$ is $0.7\text{ kcal}\cdot\text{s}^{-1}\cdot\text{m}^{-2}$. The amount of heat radiated in 20 seconds is

• A. $2.8\text{ kcal}$
• B. $0.28\text{ kcal}$
• C. $5.6\text{ kcal}$
• D. $0.56\text{ kcal}$

Hint:

When multiplying decimals like $0.7 \times 0.04 \times 20$, combine the constants first: $0.04 \times 20 = 0.8$. Then, simply calculate $7 \times 8 = 56$ and place the decimal point two positions over to get $0.56$ without any scratching!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a radiating sphere with a surface area $A = 0.04\text{ m}^2$ and an emissive power $E = 0.7\text{ kcal}\cdot\text{s}^{-1}\cdot\text{m}^{-2}$. We need to compute the total thermal energy $Q$ radiated by this sphere over a time interval of $t = 20\text{ seconds}$.

Step 2: Key Formula or Approach:
Emissive power ($E$) is defined as the total radiant energy emitted per unit surface area per unit time: $$E = \frac{Q}{A \cdot t}$$ Rearranging this definition allows us to calculate the total radiated heat energy directly: $$Q = E \cdot A \cdot t$$

Step 3: Detailed Explanation:
Substitute the given values into our energy equation: $$Q = 0.7 \times 0.04 \times 20$$ We can regroup the numbers to make the mental math simpler: $$Q = 0.7 \times (0.04 \times 20) = 0.7 \times 0.8$$ $$Q = 0.56\text{ kcal}$$

Step 4: Final Answer:
The total heat energy radiated in 20 seconds is $0.56\text{ kcal}$, which corresponds to option (D).