Question

Energy is being emitted from the surface of a black body at $127{}^\circ C$ at the rate of $(1.0\times {{10}^{6}})/s{{m}^{2}}$ . The temperature of a black body at which the rate of energy emission is $(16.0\times {{10}^{6}})/s{{m}^{2}}$ will be

• A. $727{}^\circ C$
• B. $527{}^\circ C$
• C. $508{}^\circ C$
• D. $254{}^\circ C$

Answer 1

Answer: (B)

$527{}^\circ C$

Answer 2

Here, ${{T}_{1}}=127{}^\circ C=400K$ ${{E}_{2}}=16\times {{10}^{6}}J/s{{m}^{2}}$ ${{E}_{1}}=1\times {{10}^{6}}J/s{{m}^{2}}$ Using the relation. $\frac{{{E}_{2}}}{{{E}_{1}}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{4}}$ $\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{{{E}_{2}}}{{{E}_{1}}}={{\left( \frac{16.0\times {{10}^{6}}}{1\times {{10}^{6}}} \right)}^{1/4}}=2$ ${{T}_{2}}=2\times {{T}_{1}}=2\times 400=800K$ ${{T}_{2}}=527{}^\circ C$