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Step-by-Step Solution
Step 1: Understand the Problem
We have a black body initially at a temperature of 727°C (which is 1000 K). Its radiating power under these conditions is 60 W. The surroundings are at 227°C (which is 500 K). We then increase the black body’s temperature to 1227°C (which is 1500 K) and need to find the new radiating power.
Step 2: List the Known Values
Initial temperature of black body, $T_1 = 727^\circ \text{C} = 1000\,\text{K}$
Surrounding temperature, $T_0 = 227^\circ \text{C} = 500\,\text{K}$
Initial radiating power, $E_1 = 60\,\text{W}$
New temperature of black body, $T_2 = 1227^\circ \text{C} = 1500\,\text{K}$
Required: New radiating power, $E_2$
Step 3: Apply the Stefan-Boltzmann Law
For a black body, the net radiating power is given by:
$E = \sigma \bigl(T^4 - T_0^4\bigr) \, A,$
where $ \sigma $ is the Stefan-Boltzmann constant and $A$ is the surface area of the black body.
Step 4: Express the Initial Condition
Using the above relation for the initial situation:
$E_1 = \sigma \bigl(T_1^4 - T_0^4\bigr)\,A.$
Substitute the known values:
$60 = \sigma \bigl((1000)^4 - (500)^4\bigr) A. \quad \quad (1)$
Step 5: Express the Final Condition
For the black body’s new temperature, $T_2 = 1500\,\text{K}$, the radiating power $E_2$ is:
$E_2 = \sigma \bigl(T_2^4 - T_0^4\bigr)\,A = \sigma \bigl((1500)^4 - (500)^4\bigr) A. \quad \quad (2)$
Step 6: Form a Ratio to Eliminate Unknowns
Divide equation (2) by equation (1) to cancel out $ \sigma A $:
$\displaystyle \frac{E_2}{E_1}
= \frac{\bigl((1500)^4 - (500)^4\bigr)}{\bigl((1000)^4 - (500)^4\bigr)}.$
Notice that $500 = 5 \times 100$, $1000 = 10 \times 100$, and $1500 = 15 \times 100$. Factoring out $(100)^4$ from numerator and denominator simplifies it to:
$\displaystyle \frac{E_2}{60}
= \frac{(15^4 - 5^4)}{(10^4 - 5^4)}.$
Compute each term:
$5^4 = 625, \quad 10^4 = 10000, \quad 15^4 = 50625.\!
$
Thus,
$\displaystyle \frac{E_2}{60}
= \frac{(50625 - 625)}{(10000 - 625)}
= \frac{50000}{9375}.$
Step 7: Solve for the New Radiating Power
From the ratio, we have
$E_2 = 60 \times \frac{50000}{9375} = 60 \times \frac{50000}{9375} = 320\,\text{W}.$
Step 8: Final Answer
Hence, when the black body is at 1227°C, its radiating power is 320 W.
