
A metal rod at a temperature of 150°C, radiates energy at a rate of 20 W. If its temperature is increased to 300°C, then it will radiate at the rate of?
(a) 40.8 W
(b) 17.5 W
(c) 67.3 W
(d) 37.2
Answer
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Hint: In this problem, a metal rod radiates heat when its temperature is increased, this is a common phenomenon and can be explained by using Stefan’s Boltzmann law. We get the value of heat radiated per area.
Complete step by step answer:The heat energy radiated by a hot body like metal per second per unit area is proportional to the fourth power of the temperature of that body, writing it mathematically, \[P=\sigma A{{T}^{4}}\], Here A is the area of the body and T is the absolute temperature.
Initial temperature= \[{{150}^{0}}C\]= \[(150+273)=423K\]
Final temperature= \[{{300}^{0}}C\]=\[(300+273)=573K\]
Thus, taking the ratio of power radiated in two cases we get,
\[\dfrac{{{P}_{1}}}{{{P}_{2}}}={{\{\dfrac{{{T}_{1}}}{{{T}_{2}}}\}}^{4}}={{\{\dfrac{423}{573}\}}^{4}}=0.297\]
Now final radiated energy will be \[{{P}_{2}}\]=\[\dfrac{{{P}_{1}}}{0.297}\]=\[\dfrac{20}{0.297}=67.3W\]
Thus, now the metal will radiate at the rate of 67.3 W, so, the correct option is (C)
Note:while solving such kind of problems we have to keep in mind that all temperature readings have to be put in Kelvin. But if there is change in temperature then there is no issue of converting the given unit into kelvin. Because the difference always remains the same in any units.
Complete step by step answer:The heat energy radiated by a hot body like metal per second per unit area is proportional to the fourth power of the temperature of that body, writing it mathematically, \[P=\sigma A{{T}^{4}}\], Here A is the area of the body and T is the absolute temperature.
Initial temperature= \[{{150}^{0}}C\]= \[(150+273)=423K\]
Final temperature= \[{{300}^{0}}C\]=\[(300+273)=573K\]
Thus, taking the ratio of power radiated in two cases we get,
\[\dfrac{{{P}_{1}}}{{{P}_{2}}}={{\{\dfrac{{{T}_{1}}}{{{T}_{2}}}\}}^{4}}={{\{\dfrac{423}{573}\}}^{4}}=0.297\]
Now final radiated energy will be \[{{P}_{2}}\]=\[\dfrac{{{P}_{1}}}{0.297}\]=\[\dfrac{20}{0.297}=67.3W\]
Thus, now the metal will radiate at the rate of 67.3 W, so, the correct option is (C)
Note:while solving such kind of problems we have to keep in mind that all temperature readings have to be put in Kelvin. But if there is change in temperature then there is no issue of converting the given unit into kelvin. Because the difference always remains the same in any units.
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