A black body of mass 34.38 g and surface are | vedclass

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(C) The rate of cooling of a body is given by the formula: $\frac{d	heta}{dt} = \frac{\sigma A(T^4 - T_0^4)}{ms}$.Here,$m = 34.38 	imes 10^{-3} \ kg

Vedclass · Accepted Answer

$1400$

Vedclass · Answer

(C) The rate of cooling of a body is given by the formula: $\frac{d	heta}{dt} = \frac{\sigma A(T^4 - T_0^4)}{ms}$.Here,$m = 34.38 	imes 10^{-3} \ kg$,$A = 19.2 	imes 10^{-4} \ m^2$,$T = 400 \ K$,$T_0 = 300 \ K$,and $\frac{d	heta}{dt} = 0.04 \ K/s$.Rearranging for specific heat $s$: $s = \frac{\sigma A(T^4 - T_0^4)}{m(\frac{d	heta}{dt})}$.Substituting the values:$s = \frac{(5.73 	imes 10^{-8}) 	imes (19.2 	imes 10^{-4}) 	imes (400^4 - 300^4)}{(34.38 	imes 10^{-3}) 	imes 0.04}$.$s = \frac{(5.73 	imes 10^{-8}) 	imes (19.2 	imes 10^{-4}) 	imes (256 	imes 10^8 - 81 	imes 10^8)}{34.38 	imes 10^{-3} 	imes 0.04}$.$s = \frac{5.73 	imes 19.2 	imes 10^{-12} 	imes 175 	imes 10^8}{34.38 	imes 10^{-3} 	imes 0.04}$.$s = \frac{19249.92 	imes 10^{-4}}{1.3752 	imes 10^{-3}} \approx 1400 \ J \ kg^{-1} \ K^{-1}$.

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