TS EAMCET · Physics · Thermal Properties of Matter
A slab of stone of area \(3600 \mathrm{~cm}^2\) and thickness \(10 \mathrm{~cm}\) is exposed on the lower surface to steam at \(100^{\circ} \mathrm{C}\). A block of ice at \(0^{\circ} \mathrm{C}\) rests on upper surface of the slab. In one how \(4.8 \mathrm{~kg}\) of ice is melted. The thermal conductivity of the stone in \(\int \mathrm{s}^{-1} \mathrm{~m}^{-1} \mathrm{k}^{-1}\) is (Latent heat of ice \(=3.36 \times 10^5 \mathrm{~J} / \mathrm{kg}\) )
- A 12
- B 10.5
- C 1.02
- D 1.24
Answer & Solution
Correct Answer
(D) 1.24
Step-by-step Solution
Detailed explanation
Given, area of slab of stone \((A)=3600 \mathrm{~cm}^2\) \[ =3600 \times 10^{-4} \mathrm{~m}^2 \] Thickness of stone slab \(=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}\) Temperature difference \((\Delta \theta)=100^{\circ} \mathrm{C}\) We know that,…
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