The only possibility of heat flow in a thermos flask is through its cork which is `75 cm^(2)` in area and 5 cm thick. Its thermal conductivity is `0.0075 cal//cm sec^(@)C`. How lolng will 500 g of ice at `0^(@)C` in thermos flask will is `40^(@)C` and latent heat of ise is `80 cal.//gram`.

The only possibility of heat flow in a thermos flask is through its co

1.	The only possibility of heat flow in a thermos flask is through its cork which is `75 cm^(2)` in area and 5 cm thick. Its thermal conductivity is `0.0075 cal//cm sec^(@)C`. How lolng will 500 g of ice at `0^(@)C` in thermos flask will is `40^(@)C` and latent heat of ise is `80 cal.//gram`.
Answer» Correct Answer - `[2.47 hr.]` `Q=mL =KA((Delta T)/(Delta x))(Delta t)` `500xx80=0.0075xx75xx((40-0))/(5)Delta t` On solving, `Delta t=8.9xx10^(3) s= 2.47 hr`

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The only possibility of heat flow in a thermos flask is through its cork which is `75 cm^(2)` in area and 5 cm thick. Its thermal conductivity is `0.0075 cal//cm sec^(@)C`. How lolng will 500 g of ice at `0^(@)C` in thermos flask will is `40^(@)C` and latent heat of ise is `80 cal.//gram`.

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