A metal cooking pot has a base area of 0.2 ,m | vedclass

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(B) Given: Area $A = 0.2 \,m^2$,thickness $d = 2.0 \,cm = 0.02 \,m$,thermal conductivity $K = 120 \,J s^{-1} m^{-1} K^{-1}$,latent heat $L = 2 	imes

Vedclass · Accepted Answer

$183$

Vedclass · Answer

(B) Given: Area $A = 0.2 \,m^2$,thickness $d = 2.0 \,cm = 0.02 \,m$,thermal conductivity $K = 120 \,J s^{-1} m^{-1} K^{-1}$,latent heat $L = 2 	imes 10^6 \,J/kg$,and rate of boiling $dm/dt = 3.0 \,kg/min = 3.0/60 \,kg/s = 0.05 \,kg/s$.The rate of heat transfer through the base of the pot is given by the conduction formula: $\frac{dQ}{dt} = \frac{KA(T - T_{water})}{d}$.The rate of heat required to boil water is: $\frac{dQ}{dt} = L \frac{dm}{dt}$.Equating the two: $\frac{KA}{d} (T - 100) = L \frac{dm}{dt}$.Substituting the values: $\frac{120 	imes 0.2}{0.02} (T - 100) = (2 	imes 10^6) 	imes 0.05$.$1200 (T - 100) = 100,000$.$T - 100 = \frac{100,000}{1200} = 83.33$.$T \approx 183.33^{\circ} C$. Thus,the temperature is approximately $183^{\circ} C$.

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