Respuesta :
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
Learn more about heat transfer:
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