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An ice cube of mass 0.1 kg at 0^@C is pl...

An ice cube of mass 0.1 kg at `0^@C` is placed in an isolated container which is at `227^@C`. The specific heat s of the container varies with temperature T according to the empirical relation `s=A+BT`, where `A= 100 cal//kg.K and B = 2xx 10^-2 cal//kg.K^2`. If the final temperature of the container is `27^@C`, determine the mass of the container.
(Latent heat of fusion for water = `8xx 10^4 cal//kg`, specific heat of water `=10^3 cal//kg.K`).

A

0.495 kg

B

0.595 kg

C

0.695 kg

D

0.795 kg

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An ice cube of mass 0.1kg at 0^@C is placed in an isolated container which is at 227^@C . The specific heat S of the container varies with temperature T according to the empirical relation S=A+BT , where A=100 cal//kg-K and B=2xx10^-2cal//kg-K^2 . If the final temperature of the container is 27^@C , determine the mass of the container. (Latent heat of fusion of water = 8xx10^4cal//kg , Specific heat of water= 10^3cal//kg-K ).

Let cube of mass 0.2 kg at 0^@C be placed in a container whose temperature is 127^@C . The specific heat of the container varies with temperature T as s = p+qT^2 , where p = 120 cal/kg K and q = 0.03 cal//kg K^3 . If the final temperature of the container is 27^@C , what will be its mass? Take, latent heat of fusion of water = 8 xx 10^4 cal/kg and specific heat of water = 1000 cal/kg K.

The amount of heat required to raise the temperature of 75 kg of ice at 0°C to water at 10°C is (latent heat of fusion of ice is 80 cal/g, specific heat of water is 1 cal/g °C)

1 g of steam at 100^@C and an equal mass of ice at 0^@C are mixed. The temperature of the mixture in steady state will be (latent heat of steam =540 cal//g , latent heat of ice =80 cal//g ,specific heat of water =1 cal//g^@C )

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 .

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A colorimeter contains 400 g of water at a temperature of 5^(@)C . Then, 200 g of water at a temperature of +10^(@)C and 400 g of ice at a temperature of -60^(@)C are added. What is the final temperature of the contents of calorimeter? Specific heat capacity of water -1000 cal //kg//K Specific latent heat of fusion of ice =80xx1000 cal //kg Relative specific heat of ice =0.5

50 g of ice at 0^@C is mixed with 50 g of water at 80^@C . The final temperature of the mixture is (latent heat of fusion of ice =80 cal //g , s_(w) = 1 cal //g ^@C)

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 .

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