Question

One end of a steel rod of length 1.0 m is kept in ice at ${{0}^{\circ }}C$ and the other end is kept in the boiling water at ${{100}^{\circ }}C$ . The area of the cross section of rod is $0.04c{{m}^{2}}$ . Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice $=3.36\times {{10}^{5}}Jk{{g}^{-1}}$

Answer 1

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

Complete step by step answer:We have:
\[\begin{align}
  & k=45J{{s}^{-1}}{{C}^{-1}}{{m}^{-1}} \\
 & l=1m \\
 & A=0.04{{m}^{2}}=4\times {{10}^{-6}}{{m}^{2}} \\
\end{align}\]
We know that
\[Q=\dfrac{kA({{T}_{1}}-{{T}_{2}})t}{l}\]
Now putting the values and solving we get,
\[\begin{align}
  & Q=\dfrac{46\times 4\times {{10}^{-6}}(100-0)\times 1}{1} \\
 & =184\times {{10}^{-4}}......(1)
\end{align}\]
Also, heat can be written as \[Q=mL......(2)\]
Now, substitute the heat required from equation (1) in equation (2), we get:
\[\begin{align}
  & mL=184\times {{10}^{-4}} \\
 & m=\dfrac{184\times {{10}^{-4}}}{33.6\times {{10}^{5}}} \\
 & m=5.5\times {{10}^{-5}}gm \\
\end{align}\]
Hence, the mass of ice is \[5.5\times {{10}^{-5}}gm\]

Note:The amount of heat required to convert one-unit amount of substance from the solid phase to the liquid phase — leaving the temperature of the system unaltered — is known as the latent heat of fusion.
Thermal conductivity is defined as the ability of a given material to conduct/transfer heat. It is generally denoted by the symbol 'k' but can also be denoted by 'λ' and 'K'. The reciprocal of this quantity is known as thermal resistivity.