Answer

Verified

420k+ views

**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.

**Complete step by step answer:**

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.

Recently Updated Pages

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Advantages and disadvantages of science

10 examples of friction in our daily life

Trending doubts

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

10 examples of law on inertia in our daily life

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths