How much steam at 100^@C will just melt ...

How much steam at `100^@C` will just melt 1600 g of ice at `-8^@C`? (Specific heat of ice=`0.5 (cal)/g^@C`, specific heat of water= 1/cal/g^@C`, latent of fusion of ice = `80 cal/g`, and latent heat of vaporisation of water=` 540 cal/g`)

425 g

210g

365 g

198 g

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The specific heat of ice at 0^@C is

What does the statement, 'latent heat of fusion for ice is 80 cal/g' mean?

Complete the analogy: Latent heat of fusion of ice : 80 cal/g :: steam:

If 80 g steam of temperature 97^@C is released on an ice slab of temperature 0^@C ,how much ice will melt?How much energy will be transferred to water? Given : specific heat capacity of water, c = 1 cal/g °C Latent heat of melting of ice = 80 cal/g Latent heat of vaporization of water = 540 cal/g

If 80 g steam of temperature 87^@C is released on an ice slab of temperature O^@C ,how much ice will melt?How much energy will be transferred to water? Given : specific heat capacity of water, c = 1 cal/g °C Latent heat of melting of ice = 80 cal/g Latent heat of vaporization of water = 540 cal/g

1 of a steam at 100^@C melts how much ice at 0^@C ? ( Latent heat of ice=80 cal/g and latent heat of steam =540 cal/g )

How much heat will be required to convert 20g of water at 100⁰C into steam at 100⁰C ? (Specific latent heat of vaporization of water = 540 cal/g)

M' kg of water at 't' ^@C is divided into two parts so that one part of mass 'm' kg when converted into ice 0^@C would release enough heat to vapourise the other part, then m/M is equal to [Specific heat of water = 1 calg^-1^@C^-1 Latent heat of fusion of ice= 80 calg^-1 Latent heat of steam = 540 cal g^-1 ]

The water equivalent of a calorimeter is 10 g and it contains 50 g of water at 15^@C . Some amount of ice, initially -10^@C is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice = 0.5 cal g^-1^@C^-1, specific heat of water = 1.0 cal g^1^@C^-1 and latent heat of melting of ice =80 cal g^-1 )?

30 gm of ice at 0^@C is placed in a calorimeter of mass 20 gm also at 0^@C . 100 gm of water at 90^@C is poured into the calorimeter. Find the final temperature. (specific heat of calorimeter = 0.4 cal//gm ^@C , Specific heat of water = 0.42 cal//gm ^@C Latent heat of fusion of ice = 80calg )

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