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Physics Question on thermal properties of matter

A uniform copper rod of 50 cm50\, cm length is insulated on the sides, and has its ends exposed to ice and steam respectively. If there is a layer of water 1 mm thick at each end, the temperature gradient (in ∘Cm−1^{\circ}C m^{-1}) in the bar is (Assume that the thermal conductivity of copper is 400 Wm−1 K−1400 \, Wm^{-1} \, K^{-1} and water is 0.4 Wm−1 K−10.4 \, Wm^{-1} \, K^{-1})

A

60

B

40

C

50

D

55

Answer

40

Explanation

Solution

Given, Kc=400 WM−1K−1K_{c} =400\, WM ^{-1} K ^{-1}
Kw=0.4 WM−1K−1K_{w} =0.4\, WM ^{-1} K ^{-1}
θ1=100∘C,θ2=0∘C\theta_{1}= 100^{\circ} C , \theta_{2}=0^{\circ} C

In steady state, flow of heat will be same throughout the whole system
KwA(100−θ)10−3=KcA(θ−θ′)(50/100)\frac{K_{w} A(100-\theta)}{10^{-3}}=\frac{K_{ c } A\left(\theta-\theta'\right)}{(50 / 100)}
or (100−θ)=KcKw×10−3×(θ−θ′)0.5(100-\theta) =\frac{K_{c}}{K_{w}} \times 10^{-3} \times \frac{\left(\theta-\theta'\right)}{0.5}
=4000.4×10−3×(θ−θ′)0.5=\frac{400}{0.4} \times 10^{-3} \times \frac{\left(\theta-\theta'\right)}{0.5}
(100−θ)=2(θ−θ′)(100-\theta) = 2 (\theta-\theta')
⇒(100−θ)=2(θ−100+θ)\Rightarrow (100-\theta)=2(\theta-100+\theta)
(∵100−θ=θ′\because 100-\theta=\theta' given)
or (100−θ)=4θ−200(100-\theta) =4 \theta-200
⇒ 5θ=300⇒θ=60∘C\Rightarrow\, 5 \theta=300 \Rightarrow \theta=60^{\circ} C
∴θ′=100−θ=100−60=40∘C\therefore \theta'= 100-\theta=100-60=40^{\circ} C
Hence, temperature gradient along the bar
=θ−θ′0.5=60−400.5=40∘ cm=\frac{\theta-\theta'}{0.5}=\frac{60-40}{0.5}=40^{\circ}\, cm