Question

One end of metal bar of area of cross section $5\;{\rm{c}}{{\rm{m}}^{\rm{2}}}$ and $25\;{\rm{cm}}$ in length is in steam other in contact with ice, the amount of ice melts in one minute is $\left( {{L_{ice}} = 80\;{\rm{cal/gm}},\;K = 0.8\;{\rm{cgs}}\;{\rm{units}}} \right)$

1. $16\;{\rm{gm}}$
2. $12\;{\rm{gm}}$
3. $24\;{\rm{gm}}$
4. $36\;{\rm{gm}}$

Answer 1

The above problem can be resolved using the concepts and application of the heat transfer. There are three basic modes of heat transfer; this includes conduction, convection, and radiation. The conduction mode of heat transfer is that mode, in which the thermal energy transfer takes place utilizing the molecular vibration. The formula for the conduction heat transfer is being applied in equilibrium to the heat transfer during melting. Then, the final result of mass is obtained by substituting the values in the equation.

Complete step by step answer:
Given:
The thermal conductivity is, $K = 0.8\;{\rm{cgs}}\;{\rm{unit}}$.
The latent heat of ice is, ${L_{ice}} = 80\;{\rm{cal/gm}}$.
The cross- section area of the metal bar is, $A = 5\;{\rm{c}}{{\rm{m}}^{\rm{2}}}$.
The length of the bar is, $L = 25\;{\rm{cm}}$.
The time required to melt is, $t = 1\;\min. = 1\;\min. \times \dfrac{{60\;\sec }}{{1\;\min .}} = 60\;\sec$.
The expression for the rate of heat flow is given as,
$q = \dfrac{{mL}}{t}...................................\left( 1 \right)$
The expression for the heat flow due to conduction is,
${q_1} = KA\left( {\dfrac{{dT}}{L}} \right).....................................\left( 2 \right)$
Here, dT is the temperature of the formation of steam and its value is $dT = 100\;{\rm{^\circ C}}$.
Comparing the expressions of equation 1 and 2 as,

q = {q_1}\\\ \dfrac{{mL}}{t} = KA\left( {\dfrac{{dT}}{L}} \right) \end{array}$$ Solve by substituting the values in above equation as, $$\begin{array}{l} \dfrac{{m{L_{ice}}}}{t} = KA\left( {\dfrac{{dT}}{L}} \right)\\\ m = \dfrac{{\left( {KA\dfrac{{dT}}{L}} \right) \times t}}{{{L_{ice}}}}\\\ m = \dfrac{{\left( {0.8\;{\rm{cgs}} \times 5\;{\rm{c}}{{\rm{m}}^{\rm{2}}} \times \dfrac{{100\;^\circ {\rm{C}}}}{{25\;{\rm{cm}}}}} \right) \times 60\;\sec }}{{80\;{\rm{cal/gm}}}}\\\ m = 12\;{\rm{gm}} \end{array}$$ Therefore, the amount of ice that melts in 1 minute is of $$12\;{\rm{gm}}$$ and option (2) is correct. **Note:** To solve the given problem, it is necessary to understand the concepts of heat transfer. Moreover, it is also required to remember the fundamental mode of heat transfer. These include the conduction, convection, and the radiation. Along with this, these concepts are vital to remember.