Search any question & find its solution
Question:
Answered & Verified by Expert
If a $4 \mathrm{~kg}$ of ice is inside a closed cubical thermocol box of side length $20 \mathrm{~cm}$ and wall thickness $4 \mathrm{~cm}$ then the mass of the ice remaining after 10 hours is nearly
(The out side temperature $=50^{\circ} \mathrm{C}$
co-efficient of thermal conductivity of thermocol $=0.01 \mathrm{Js}^{-1} \mathrm{~m}^{-1{ }^{\circ}} \mathrm{C}^{-1}$
Latent heat of fusion of ice $=335 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1}$)
Options:
(The out side temperature $=50^{\circ} \mathrm{C}$
co-efficient of thermal conductivity of thermocol $=0.01 \mathrm{Js}^{-1} \mathrm{~m}^{-1{ }^{\circ}} \mathrm{C}^{-1}$
Latent heat of fusion of ice $=335 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1}$)
Solution:
2209 Upvotes
Verified Answer
The correct answer is:
$3.678 \mathrm{~kg}$
No solution. Refer to answer key.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.