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Freezer is at -18 deg C. Will will be Radiative heat transfer if meat is at 298k, its emissivity is 0.82 and average area is 0.045m2. Either use Stefan Boltzmann law or take hr (radiative heat transfer co-efficient) as 0.227. ε = ( Tav/100 )3

Freezer is at -18 deg C. Will will be Radiative heat transfer if meat is at 298k, its emissivity is 0.82 and average area is 0.045m2. Either use Stefan Boltzmann law or take hr (radiative heat transfer co-efficient) as 0.227. ε = ( Tav/100 )3 Correct Answer 7.61

Stefan Boltzmann law: T1 = 298, T2 = 255. σ= 5.67*10-8 W /m2K4 T14 = 7.88 * 109 T24 = 4.22* 109 Using, q = A εσ (T14 – T24) q = 5.7 * 10-8 * 0.82* 0.45 * (7.88-4.22) * 109 = 7.61. Tav= (298 + 255) /2 hr = 3.93 W/m2 K q = A εσ (T1 -T2) = 3.93* 0.045 * (298-255) = 7.61.

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