Part A One end of an insulated metal rod is maintained at 100 C and the other en

Question

Part A One end of an insulated metal rod is maintained at 100 C and the other end is maintained at 0.00 C by an ice–water mixture. The rod has a length of 55.0 cm and a cross-sectional area of 1.40 cm2 . The heat conducted by the rod melts a mass of 7.05 g of ice in a time of 15.0 min. Find the thermal conductivity k of the metal.

Part B A pot with a steel bottom 9.00

mm thick rests on a hot stove. The area of the bottom of the pot is 0.150 m2 . The water inside the pot is at 100.0 C, and 0.360 kg are evaporated every 4.00 min. Find the temperature of the lower surface of the pot, which is in contact with the stove.min.

I need help with these last two for my homework Please!!

Explanation / Answer

A)

let,

length of the rod l=55 cm

cross-sectional area A=1.4 cm^2

mass of the ice m=7.05 g

time taken to melts the ice m=15 min

and

T1=100 oC, T2=0 oC

rate of energy transferred P=Q/t=k*A*(dT)/l

Q/t=k*A*(dT)/l

(m*L_ice)/t=k*A*(dT)/l

(7.05*10^-3*3.33*10^5)/(15*60)=k*(1.4*10^-4)*(100)/(55*10^-2)

====> k=102.48 w/m.K

thermal conductivity of rod k=102.48 w/m.K

B)

let,

thickness of the steel l=9 mm

area A=0.15 m^2

tempetarture inside the pot is T1=100 oC

tempetarture lowersurface the pot is T2

water mass m=0.36

time taken to evaporated t=4 min

and

thermal conductivity of steel k=50.2 w/m.K

rate of energy lost is,

Q/t=k*A*(dT)/l

(m*L_v)/t=k*A*(dT)/l

(0.36*22.6*10^5)/(4*60)=(50.2)*(0.15)*(T2-100)/((9*10^-3)

====> T2=104.052 oC

tempetarture lowersurface the pot is T2=104.052 oC

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