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veemo
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[SOLVED] Thermal physics problem
A steel bar 10cm long is welded end-to-end to a copper bar 20cm long. Both bars are insulated perfectly along their sides. Each bar has a radius of 2.0cm. The free end of the steel bar is maintained at 100^{o}C and the free end of the copper bar is maintained at 0^{o}C. Find the temperature at the junction between the two bars and the total rate of flow of heat.
k_{steel}=50.2Wm^{-1}K^{-1}
k_{copper}=385.0Wm^{-1}K^{-1}
H = \frac{dQ}{dt} = -kA\frac{dT}{dx}
Copper:
\frac{dQ}{dt} = \frac{-kA(373-T)}{L} = \frac{-385\pi(2X10^{-2})^{2}(273-T)}{20X10^{-2}}
Where T is the point at the junction between the bars, piXr^{2} has been substituted in for A, and 273K = 0C
so \frac{dQ}{dt} = -2.42(273-T)Steel:
\frac{dQ}{dt} = \frac{-50.2\pi(2X10^{-2})^{2}(373-T)}{10X10^{-2}}
As 373K = 100C
so \frac{dQ}{dt} = -0.63(373-T)
I then said H_{copper} = H_{steel} which I'm not sure is correct as in the example I was given in class the two bars were of equal length, but continued with the following:
2.42(273-T)=0.63(373-T)
2.42T-0.63T=660.4-235.3
T=\frac{425.1}{1.79}= 238K
I'm not sure this is right as it is less than the temperature of the free end of the copper bar and surely the result should be between 273K and 373K?
Also I don't know where to start on finding the total rate of flow of heat... The only thing I could think of was adding the individual heat flows together but I'm sure there's more to it than that.
Homework Statement
A steel bar 10cm long is welded end-to-end to a copper bar 20cm long. Both bars are insulated perfectly along their sides. Each bar has a radius of 2.0cm. The free end of the steel bar is maintained at 100^{o}C and the free end of the copper bar is maintained at 0^{o}C. Find the temperature at the junction between the two bars and the total rate of flow of heat.
k_{steel}=50.2Wm^{-1}K^{-1}
k_{copper}=385.0Wm^{-1}K^{-1}
Homework Equations
H = \frac{dQ}{dt} = -kA\frac{dT}{dx}
The Attempt at a Solution
Copper:
\frac{dQ}{dt} = \frac{-kA(373-T)}{L} = \frac{-385\pi(2X10^{-2})^{2}(273-T)}{20X10^{-2}}
Where T is the point at the junction between the bars, piXr^{2} has been substituted in for A, and 273K = 0C
so \frac{dQ}{dt} = -2.42(273-T)Steel:
\frac{dQ}{dt} = \frac{-50.2\pi(2X10^{-2})^{2}(373-T)}{10X10^{-2}}
As 373K = 100C
so \frac{dQ}{dt} = -0.63(373-T)
I then said H_{copper} = H_{steel} which I'm not sure is correct as in the example I was given in class the two bars were of equal length, but continued with the following:
2.42(273-T)=0.63(373-T)
2.42T-0.63T=660.4-235.3
T=\frac{425.1}{1.79}= 238K
I'm not sure this is right as it is less than the temperature of the free end of the copper bar and surely the result should be between 273K and 373K?
Also I don't know where to start on finding the total rate of flow of heat... The only thing I could think of was adding the individual heat flows together but I'm sure there's more to it than that.
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