Question

Compare the rate of loss of heat from a metal sphere at 827°C with the rate of loss of heat of the same sphere at 427°C if the temp of the surrounding is 27°C​

Answer 1

Answer:

kk my tq for points tq my i am playing free fire my id is alokj2508 and BOYSop123123

tq

Answer 2

Answer:

Case 1:

Temperature of metal sphere, $T_1 = 827^\circ C = 827+273 = 1100 K$

Temperature of the surrounding, $T = 27^\circ C = 27+273 = 300 K$

Rate of loss of heat,

$R_1 = (\frac{dQ}{dt})_1 \\ R_1 = e\sigma A(T_1^4 - T^4)\\where\hspace{1mm}e=emissivity\\\sigma=Stefan's\hspace{1mm}constant\\T_1 = Temperature\hspace{1mm}of\hspace{1mm}metal\hspace{1mm}sphere\\T = Temperature\hspace{1mm}of\hspace{1mm}surrounding\\A = Area$

Case 2:

Temperature of metal sphere, $T_2= 427^\circ C = 427+273 = 700 K$

Temperature of the surrounding, $T = 27^\circ C = 27+273 = 300 K$

Rate of loss of heat,

$R_2 = (\frac{dQ}{dt})_2 \\ R_2 = e\sigma A(T_2^4 - T^4)\\where\hspace{1mm}e=emissivity\\\sigma=Stefan's\hspace{1mm}constant\\T_2 = Temperature\hspace{1mm}of\hspace{1mm}metal\hspace{1mm}sphere\\T = Temperature\hspace{1mm}of\hspace{1mm}surrounding\\A = Area$

Dividing R₁ by R₂, we get

$\frac{R_1}{R_2} = \frac{e\sigma A(T_1^4-T^4)}{e\sigma A(T_2^4-T^4)}$

$\frac{R_1}{R_2} = \frac{(T_1^4-T^4)}{(T_2^4-T^4)}$

$\frac{R_1}{R_2} = \frac{(1100)^4 - (300)^4}{(700)^4-(300)^4}$

$\frac{R_1}{R_2} = \frac{1456}{232} = \frac{182}{29}$

R₁ : R₂ = 182 : 29

Hence, the ratio of rate of heat loss is 182 : 29.

#SPJ3