Two black bodies A and B have equal surface areas and are maintained at temperatures 27°C and 177°C respectively. What will be the ratio of the thermal energy radiated per second by A to that by B?

Question

Two black bodies A and B have equal surface areas and are maintained at temperatures 27°C and 177°C respectively. What will be the ratio of the thermal energy radiated per second by A to that by B? (A) 4:9 (B) 2:3 (C) 16:81 (D) 27:177

Tardigrade Admin · Accepted Answer

Q=σ AT4, (Q1/Q2) =((T1/T2))4 =((273+27/273+177))4 =((300/450))4 =((2/3))4 =(16/81)

Q. Two black bodies A and B have equal surface areas and are maintained at temperatures 27°C and 177°C respectively. What will be the ratio of the thermal energy radiated per second by A to that by B?

4:9

2:3

16:81

27:177

$Q = σ A T^{4}, \frac{Q _{1}}{Q _{2}} = (\frac{T _{1}}{T _{2}})^{4} = (\frac{273 + 27}{273 + 177})^{4} = (\frac{300}{450})^{4} = (\frac{2}{3})^{4} = \frac{16}{81}$

Q. Two black bodies A and B have equal surface areas and are maintained at temperatures 27°C and 177°C respectively. What will be the ratio of the thermal energy radiated per second by A to that by B?

4:9

2:3

16:81

27:177

Q=σAT4,Q2​Q1​​=(T2​T1​​)4=(273+177273+27​)4=(450300​)4=(32​)4=8116​

$Q = σ A T^{4}, \frac{Q _{1}}{Q _{2}} = (\frac{T _{1}}{T _{2}})^{4} = (\frac{273 + 27}{273 + 177})^{4} = (\frac{300}{450})^{4} = (\frac{2}{3})^{4} = \frac{16}{81}$