TS EAMCET · Physics · Thermal Properties of Matter
A rectangular ice box of total surface area of \(1000 \mathrm{~cm}^2\) initially contains 1.5 kg of ice at \(0{ }^{\circ} \mathrm{C}\). If the thickness of the walls of the box is 2 mm and the temperature outside the box is \(42^{\circ} \mathrm{C}\), then the mass of the ice remaining in the box after 160 minutes is (Thermal conductivity of the material of the box \(=10^{-2} \mathrm{Wm}^{-1} \mathrm{~K}^{-1}\) and latent heat of the fusion of ice \(=336 \times 10^3 \mathrm{Jkg}^{-1}\) )
- A 0.6 kg
- B 0.9 kg
- C 0.8 kg
- D 0.7 kg
Answer & Solution
Correct Answer
(B) 0.9 kg
Step-by-step Solution
Detailed explanation
\(P = \frac{kA\Delta T}{L} = \frac{10^{-2} \times (1000 \times 10^{-4}) \times 42}{2 \times 10^{-3}} = 21 \mathrm{~W}\) \(Q = P \times t = 21 \times (160 \times 60) = 201600 \mathrm{~J}\) \(m_{melted} = \frac{Q}{L_f} = \frac{201600}{336 \times 10^3} = 0.6 \mathrm{~kg}\)…
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