Question 915: Thermodynamics & KTG

Question

The emissive power of sphere of area $$0.04 \mathrm{~m}^2$$ is $$0.7 \mathrm{~k} \mathrm{~cal} \mathrm{~s}^{-1} \mathrm{~m}^{-2}$$. The amount of heat radiated in 20 second is

Accepted Answer

The amount of heat radiated by an object is given by the formula:
$$Q = E 	imes A 	imes t$$
where:
$Q$ is the amount of heat radiated,
$E$ is the emissive power (in $kcal \ s^{-1} \ m^{-2}$),
$A$ is the surface area (in $m^2$), and
$t$ is the time in seconds.
Given that:
Emissive power, $E = 0.7 \ kcal \ s^{-1} \ m^{-2}$
Area, $A = 0.04 \ m^2$
Time, $t = 20 \ s$
Substituting the given values into the formula gives:
$$Q = 0.7 	imes 0.04 	imes 20$$
$$Q = 0.56 \ kcal$$
Hence, the amount of heat radiated in 20 seconds is $0.56 \ k \mathrm{cal}$, which corresponds to Option D.

Answer

$$2.8 \mathrm{~k} \mathrm{~cal} \mathrm{~s}^{-1} \mathrm{~m}^{-2}$$

Answer

$$0.28 \mathrm{~k} \mathrm{~cal} \mathrm{~s}^{-1} \mathrm{~m}^{-2}$$

Answer

$$5.6 \mathrm{~k} \mathrm{~cal} \mathrm{~s}^{-1} \mathrm{~m}^{-2}$$

Answer

$$0.56 \mathrm{~k} \mathrm{~cal} \mathrm{~s}^{-1} \mathrm{~m}^{-2}$$

Thermodynamics & KTG

The emissive power of sphere of area $0.04 \mathrm{~m}^2 is 0.7 \mathrm{~k} \mathrm{~cal} \mathrm{~s}^{-1} \mathrm{~m}^{-2}$. The amount of heat radiated in 20 second is

Choose the correct answer: