Two rods, one made of brass and the other made of copper, are joined end to end.

Question

Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.300 m and the length of the copper section is 0.800 m. Each segment has cross-sectional area 0.00500 m^2. The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice-water mixture, in both cases under normal atmospheric presume. The sides of the rods are insulated so there is no heat loss to the surroundings, (a) What is the temperature of the point where the brass and copper segments are joined? (b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?

Explanation / Answer

rate of heat trnsfred by copper is

Pcu = Kcu X A X ( T - Tcu ) / Lcu

same way for brass

Pbr =Kbr X A X ( Tbr - T ) / Lbr    

Pcu = Pbr                   

Kcu X A X ( T - Tcu ) / Lcu = Kbr X A X ( Tbr - T ) / Lbr    

then to get T value

T = ( Kcu X Tcu X Lbr + Kbr X Tbr X Lcu ) / ( Kcu X Lbr + Kbr X Lcu )

by substituting the values

T= ( 385.0 X 273.15 X 0.300 + 109 X 373.15 X 0.800 ) / ( 385 X 0.300 + 109 X 0.800 )

T = 64087.505 / 202.7

= 316.16 K

= 316.16 - 273.15

T = 43.010C

P = A ( Tbr - Tcu ) / ) ( Lbr / Kbr + Lcu / Kcu )

P = 0.005 X ( 373.15 - 273.15 ) / ( 0.300 / 109 + 0.800 / 385 )

P = 1.03.511 J/sec

using equation for P

P = Q / t

Q = P X t

Q = 103.511 X 300

Q = 31053.3 J

and we know even that formula can be used ie is here

Q = m X L

then to get mass m

m = Q / L

= 31053.3 / 334.00

m = 92.973 grams of ice melt

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