A blacksmith is flattening a steel plate having dimensions $10 \mathrm {~cm} \times 15 \mathrm {~cm} \times 1 \mathrm {~mm}$
He has heated the plate to 900 K. If the emissivity of the plate is 0.75 , at what rate does it lose energy by radiation? Ignore any heat exchange with the surroundings. $\left( \sigma = 5.67 \times 10 ^ { - 8 } \mathrm {~W} / \mathrm { m } ^ { 2 } \cdot \mathrm { K } ^ { 4 } \right)$

Question

A blacksmith is flattening a steel plate having dimensions $10 \mathrm {~cm} \times 15 \mathrm {~cm} \times 1 \mathrm {~mm}$
He has heated the plate to  900 K. If the emissivity of the plate is  0.75 , at what rate does it lose energy by radiation? Ignore any heat exchange with the surroundings. $\left( \sigma = 5.67 \times 10 ^ { - 8 } \mathrm {~W} / \mathrm { m } ^ { 2 } \cdot \mathrm { K } ^ { 4 } \right)$

Quizplus · Accepted Answer

The rate at which the plate loses energy by radiation is given by the Stefan-Boltzmann law:$$P = \epsilon \sigma A T^4$$where:* $P$ is the power radiated (in watts)* $\epsilon$ is the emissivity of the plate (0.75)* $\sigma$ is the Stefan-Boltzmann constant $(5.67 \times 10^{-8} \mathrm{~W/m^2 \cdot K^4})$* $A$ is the surface area of the plate $(0.1 \mathrm{~m} \times 0.15 \mathrm{~m} = 0.015 \mathrm{~m^2})$* $T$ is the temperature of the plate (900 K)Plugging in these values, we get:$$P = 0.75 \times 5.67 \times 10^{-8} \mathrm{~W/m^2 \cdot K^4} \times 0.015 \mathrm{~m^2} \times (900 \mathrm{~K})^4 = 850 \mathrm{~W}$$Therefore, the correct answer is A: 850 W.

A Blacksmith Is Flattening a Steel Plate Having Dimensions 10 cm×15 cm×1 mm10 \mathrm {~cm} \times 15 \mathrm {~cm} \times 1 \mathrm {~mm}10 cm×15 cm×1 mm

A Blacksmith Is Flattening a Steel Plate Having Dimensions $10 \mathrm {~cm} \times 15 \mathrm {~cm} \times 1 \mathrm {~mm}$