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Question: One end of a 0.25 m long metal bar is in steam and the other is in contact with ice. If 12 g of ice ...

One end of a 0.25 m long metal bar is in steam and the other is in contact with ice. If 12 g of ice melts per minute, then the thermal conductivity of the metal is (Given: cross section of the bar is 5×10−4m25\times {{10}^{-4}}{{m}^{2}} and latent heat of ice is 80calg−180cal{{g}^{-1}} )
a)20cal s−1m−1∘C−120cal\text{ }{{s}^{-1}}{{m}^{-1}}^{\circ }{{C}^{-1}}
b)10cal s−1m−1∘C−110cal\text{ }{{s}^{-1}}{{m}^{-1}}^{\circ }{{C}^{-1}}
c)40cal s−1m−1∘C−140cal\text{ }{{s}^{-1}}{{m}^{-1}}^{\circ }{{C}^{-1}}
d)30cal s−1m−1∘C−130cal\text{ }{{s}^{-1}}{{m}^{-1}}^{\circ }{{C}^{-1}}

Explanation

Solution

As we know that, the amount of heat required is given by Q=mLQ=mL, where Q is heat, m is mass and L is latent heat. Find the amount of heat required by using this formula. Also, heat can be written as Q=KA(T1−T2)txQ=\dfrac{KA({{T}_{1}}-{{T}_{2}})t}{x}, where K is thermal conductivity coefficient, A is cross-section area, (T1−T2)\left( {{T}_{1}}-{{T}_{2}} \right) is change of temperature, t is time taken and x is length of rod. Now, substitute the heat required in this formula and find the value of thermal conductivity of the rod.

Complete step by step answer: We have:
Temperature of ice is 0∘C{{0}^{\circ }}C
Temperature of steam is 100∘C{{100}^{\circ }}C
So, the change of temperature across the ends of the rod is T1−T2=100−0=100∘C{{T}_{1}}-{{T}_{2}}=100-0={{100}^{\circ }}C
The time taken to melt the ice is t=1min⁡=60sec⁡t=1\min =60\sec
Mass of ice = 12g
The length of the metal bar is x=0.25mx=0.25m
The cross-section area of rod is A=5×10−4m2A=5\times {{10}^{-4}}{{m}^{2}}
A=5×10−4m2,L=80calg⁡−1A=5\times {{10}^{-4}}{{m}^{2}},L=80ca{{\lg }^{-1}}
As we know that, the amount of heat required by the ice is: Q=mLQ=mL
So, we get: Q=12×80=960calQ=12\times 80=960cal
But, Q=KA(T1−T2)txQ=\dfrac{KA({{T}_{1}}-{{T}_{2}})t}{x}
So, by comparing both the equations, we get:
960=K×5×10−4×100×600.25960=\dfrac{K\times 5\times {{10}^{-4}}\times 100\times 60}{0.25}
∴K=960×0.255×10−4×100×60\therefore K=\dfrac{960\times 0.25}{5\times {{10}^{-4}}\times 100\times 60}
K=30cals−1m−1∘C−1K=30cal{{s}^{-1}}{{m}^{-1}}^{\circ }{{C}^{-1}}

Note: a) Thermal conductivity is defined as the ability of a given material to conduct/transfer heat.
b) It is generally denoted by the symbol 'k' but can also be denoted by 'λ' and 'κ'.
c) The reciprocal of this quantity is known as thermal resistivity.