# Water is being boiled in flat bottom kettle placed on a stove. The area of the bottom is 3000cm^2 and the thickness is 2 mm. If the amount of steam

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Water is being boiled in flat bottom kettle placed on a stove. The area of the bottom is 3000cm^2 and the thickness is 2 mm. If the amount of steam produced is 1g//min, calculate the difference of temperature between the inner and outer surface of the bottom. K for the material of kettle is 0.5cal^@C//s//cm, and the latent heat of steam is 540 cal//g.
A. 0.12 xx 10^(-5)K
B. 1.9 xx 10^(-3)K
C. 1.3 xx 10^(-4^(@))C
D. 1.2 xx 10^(-3)K

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d) Rate of production of mass of steam = (dm)/(dt) = 1/(60) gs^(-1)
Rate of heat transformed by stove,
(dtheta)/(dt) = L(dm)/(dt)=540 xx 1/(60) = 9cal^(@)C^(-1)s^(-1)cm^(-1)
Also, (d(theta))/(dt) = (KAd(theta))/(dx)rArr L(dm)/(dt)=(KAd(theta))/dx
rArr 9 = (0.5 xx 3000xx d(theta))/(0.2)
rArr d(theta) = 1.2 xx 10^(-3^(@))C=1.2xx10^(-3)K