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The only possibility of heat flow in a t...


The only possibility of heat flow in a thermos flask is through its cork which is `75cm^2` in area and 5 cm thick its thermal conductivity is `0.0075 cal//cm-s-^@C`. The outside temperature is `40^@C` and latent heat of ice is `80 cal//g`. Time taken by 500 g of ice at `0^@C` in the flask to melt into water at `0^@C` is `

A

2.47 hr

B

4.27 hr

C

7.42 hr

D

4.72 hr

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The correct Answer is:
A
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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 .

The entropy change in melting 1g of ice at 0^@C . (Latent heat of ice is 80 cal g^(-1) )

The water of mass 75 g at 100^(@)C is added to ice of mass 20 g at -15^(@)C . What is the resulting temperature. Latent heat of ice = 80 cal//g and specific heat of ice = 0.5 .

Calculate change in entropy when 5g of pure ice melts to form water at 0^(@)C Given latent heat of ice is 80cal//g at 0^(@)C .

Find the heat required to convert 20 g of iceat 0^(@)C into water at the same temperature. ( Specific latent heat of fusion of ice =80 cal//g )

20 g of steam at 100^@ C is passes into 100 g of ice at 0^@C . Find the resultant temperature if latent heat if steam is 540 cal//g ,latent heat of ice is 80 cal//g and specific heat of water is 1 cal//g^@C .

What amount of ice will remains when 52 g ice is added to 100 g of water at 40^(@)C ? Specific heat of water is 1 cal/g and latent heat of fusion of ice is 80 cal/g.

50 g of steam at 100^@ C is passed into 250 g of at 0^@ C . Find the resultant temperature (if latent heat of steam is 540 cal//g , latent heat of ice is 80 cal//g and specific heat of water is 1 cal//g -.^@ C ).

There is ice formation on a tank of water of thickness 10 cm.How much time it will take to have a layer of 0.1 cm below it ?The outer temperature is -5^(@)C , the thermal conductivity of ice is 0.005 calcm^(-1) ""^(@)C^(-1) and latent heat of ice is 80cal//g and the density of ice is 0.91 gcm^(-3) :

150 g of ice is mixed with 100 g of water at temperature 80^(@)C . The latent heat of ice is 80 cal/g and the specific heat of water is 1 cal//g-.^(@)C . Assuming no heat loss to the environment, the amount of ice which does not melt is –

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