Question

A metallic ball of surface area \[0.02\,{{\text{m}}^2}\] and at a temperature of \[527^\circ {\text{C}}\] is placed in a constant temperature enclosure maintained at \[27^\circ C\]. If the absorption coefficient of metal surface is 0.4, find the rate at which heat is lost by ball \[\sigma = 5.7 \times {10^{ - 8}}\] S.I units.

Answer 1

Hint:Stefan’s law states that, as the temperature of the material decreases, it releases the heat in the form of radiation. Recall the formula for heat loss using Stefan’s law. Heat loss should be expressed as heat released per second.

Formula used:
\[\dfrac{Q}{t} = e\sigma A\left( {T_1^4 - T_2^4} \right)\]
Here, Q is the heat, t is the time, e is the emissivity, A is the surface area, \[{T_1}\] is the initial temperature, \[{T_2}\] is the final temperature and \[\sigma \] is the Stefan’s constant.

Complete step by step answer:
We have given that the surface area of the metallic ball is \[A = 0.02\,{{\text{m}}^2}\] and its initial temperature is \[{T_1} = 527^\circ {\text{C}} = 800\,{\text{K}}\]. Then the metallic ball is placed in the enclosure whose surrounding temperature is \[{T_2} = 27^\circ C = 300\,{\text{K}}\]. The coefficient of absorption or emissivity of the metallic ball is \[e = 0.4\].

We know that from Stefan’s law, as the temperature of the material decreases, it releases the heat in the form of radiation. Let us express the rate of heat loss by the metallic ball using Stefan’s law as follows,
\[\dfrac{Q}{t} = e\sigma A\left( {T_1^4 - T_2^4} \right)\]
Here, Q is the heat, t is the time \[\sigma \] is the Stefan’s constant.

Substituting \[e = 0.4\], \[\sigma = 5.7 \times {10^{ - 8}}\], \[A = 0.02\,{{\text{m}}^2}\], \[{T_1} = 800\,{\text{K}}\] and \[{T_2} = 300\,{\text{K}}\] in the above equation, we get,
\[\dfrac{Q}{t} = \left( {0.4} \right)\left( {5.7 \times {{10}^{ - 8}}} \right)\left( {0.02} \right)\left( {{{800}^4} - {{300}^4}} \right)\]
\[ \Rightarrow \dfrac{Q}{t} = \left( {0.4} \right)\left( {5.7 \times {{10}^{ - 8}}} \right)\left( {0.02} \right)\left( {4.015 \times {{10}^{11}}} \right)\]
\[ \Rightarrow \dfrac{Q}{t} = \left( {4.56 \times {{10}^{ - 10}}} \right)\left( {4.015 \times {{10}^{11}}} \right)\]
\[ \therefore \dfrac{Q}{t} = 183\,{\text{J/s}}\]

Therefore, the rate of loss of heat by the metallic ball is 183 joule per second.

Note:Stefan’s law is actually expressed as \[\dfrac{Q}{t} = e\sigma A\left( {T_2^4 - T_1^4} \right)\], that is the initial temperature is subtracted from the final temperature. If the temperature of the body decreases, the heat is released and the rate of heat loss attains a negative sign. If the heat loss has a positive sign, it implies that the heat is gained by the material. This formula is valid only when the heat is transferred in the form of radiation.