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# A slab of stone area $3600\,{\text{c}}{{\text{m}}^{\text{2}}}$ and thickness 10 cm is exposed on the lower surface to steam at $100^\circ C$. A block of ice at $0^\circ C$ rests on the upper surface of the slab. In one minute, 4.8 kg of ice is melted. The thermal conductivity of the stone is $J{s^{ - 1}}{m^{ - 1}}{k^{ - 1}}$ is. (latent heat of ice =$3.36 \times {10^5}\,J/kg$)

Last updated date: 12th Aug 2024
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Hint: The heat transferred by the steam from the lower slab is equal to the heat gained by the block of ice. Use the formula for heat transfer from the surface of thermal conductivity K of thickness L to determine the heat transferred to the ice.

Formula used:
$\dfrac{Q}{t} = \dfrac{{KA\left( {{T_2} - {T_1}} \right)}}{L}$
Here, K is the thermal conductivity, A is the cross-sectional area, ${T_2}$ and ${T_1}$ is the temperatures of the respective surfaces and L is the thickness.

The rate of heat transferred through the lower surface of the slab to the upper surface of the slab is,
$\dfrac{Q}{t} = \dfrac{{KA\left( {{T_2} - {T_1}} \right)}}{L}$
Here, K is the thermal conductivity of the stone, A is the cross-sectional area of the stone, ${T_2}$ is the temperature of the steam, ${T_1}$ is the temperature of the block of ice and L is the thickness of the slab of the stone.
Express the heat transferred to the block of ice in time t as follows,
$Q = \dfrac{{KA\left( {{T_2} - {T_1}} \right)}}{L}t$
The heat gained by the block of ice for melting (latent heat of fusion) is given as,
$Q = {m_{ice}}{L_f}$
Here, ${m_{ice}}$ is the mass of the ice and ${L_f}$ is the latent heat of fusion of ice.
Since the latent heat for the melting of the ice is the same as the heat transferred to the block of ice from the lower surface of the slab, we can write,
$\dfrac{{KA\left( {{T_2} - {T_1}} \right)}}{L}t = {m_{ice}}{L_f}$
$\Rightarrow K = \dfrac{{{m_{ice}}{L_f}L}}{{A\left( {{T_2} - {T_1}} \right)t}}$
Substitute 4.8 kg for ${m_{ice}}$, $3.36 \times {10^5}\,J/kg$ for ${L_f}$, 0.1 m for $L$, $0.36\,{{\text{m}}^{\text{2}}}$ for A, 1 s for t, $100^\circ C$ for ${T_2}$ and $0^\circ C$ for ${T_1}$ in the above equation.
$K = \dfrac{{\left( {4.8\,kg} \right)\left( {3.36 \times {{10}^5}\,J/kg} \right)\left( {0.1\,m} \right)}}{{\left( {0.36\,{m^2}} \right)\left( {\left( {100 + 273\,k} \right) - \left( {0 + 273\,k} \right)} \right)\left( {1\,s} \right)}}$
$\Rightarrow K = \dfrac{{1.6128 \times {{10}^5}\,J\,m}}{{36\,{m^2}k\,s}}$
$K = 4480\,J{m^{ - 1}}{s^{ - 1}}{k^{ - 1}}$
Therefore, the thermal conductivity of the stone is $4480\,J{m^{ - 1}}{s^{ - 1}}{k^{ - 1}}$.

Note:
In the above calculation, 273 k is the temperature in Kelvin and it's not the thermal conductivity. The temperature is given in Celsius, therefore, do convert it in Kelvin by adding 273 K in the given temperature.