Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.400 m and the length of the copper section is 0.800 m . Each segment has cross-sectional area 0.00700 m2 . 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 pressure. 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?

Answer:

a) 36°

b) 0.109 kg

Explanation:

Heat flows from brass to copper with the brass having its temperature

Length of brass = 0.4

Length of copper = 0.8

Temperature of = 36.15

See attachment for calculation

The temperature at the joint is 36.15°C

The amount of ice melted is 1.086 kg

The rate of transfer of thermal energy,

   H = Q/t = KAΔT/L

  where, K is the thermal conductivity of the substance, A is cross-sectional area, ΔT is temperature difference at the ends and L is the length

As given in the question,

  the length of the brass section   [tex]L_{1}[/tex] = 0.4 m

  it's thermal conductivity      [tex]K_{b}[/tex] = 109 J[tex]s^{-1}m^{-1}K^{-1}[/tex]

  the temperature at the brass end  [tex]T_{1}[/tex] = 373K

  the length of the copper section [tex]L_{2}[/tex] = 0.8 m

  it's thermal conductivity      [tex]K_{c}[/tex] = 385 J[tex]s^{-1}m^{-1}K^{-1}[/tex]

  the temperature at the brass end  [tex]T_{1}[/tex] = 273K

 cross-sectional area of both the substance is same A = 0.007 [tex]m^{2}[/tex]

  Let the temperature at the joint be T

The rate of heat flow must be constant across the whole length of the setup.

Hence at the joint,

   [tex]\frac{K_{b}A(T_{1}-T) }{L_{1} } =\frac{K_{c}A(T-T_{2}) }{L_{2} }[/tex]

⇒ [tex]\frac{ 109*A*(373-T)}{0.4} =\frac{385*A(T-273)}{0.8}[/tex] ⇒ T=309.15 K

⇒ T = 36.15°C is the temperature at the joint.

Now we have to calculate the equivalent thermal conductivity K of the setup in order to calculate the amount of heat transfer.

considering equivalent thermal conductivity K throughout the setup we can form the following equation to calculate its value

   [tex]\frac{KA(T_{1}-T_{2} ) }{L_{1}+L_{2} } =\frac{K_{c}A(T-T_{2}) }{L_{2} }[/tex]

   ⇒   [tex]\frac{ K*A*(100)}{1.2} =\frac{385*A(36.15)}{0.8}[/tex]

   ⇒  K = 208.76 J[tex]s^{-1}m^{-1}K^{-1}[/tex]

the amount of heat transferred at the copper end in ice-water mixture in 5 minutes(300 seconds) :

        Q = [tex]\frac{KA(T_{1}-T_{2} ) }{L_{1}+L_{2} }[/tex] × t =  [tex]\frac{208.76*0.007*100}{1.2}[/tex] × 300 = 36533 J

        latent heat of fusion of ice [tex]L_{f}[/tex] = 33600 J/kg

         [tex]Q=mL_{f}[/tex]

         [tex]m=\frac{Q}{L_{f} }[/tex]

         [tex]m=\frac{36533}{33600}[/tex] ⇒ m = 1.086 kg of ice is melted in 5 minutes

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