Home
Class 11
PHYSICS
50 g of steam at 100^@ C is passed into ...

`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`).

A

`40^@ C`

B

`30^@ C`

C

`20^@ C`

D

`10^@ C`

Text Solution

Verified by Experts

The correct Answer is:
A
Heat lost by steam = Heat gained by ice.
`m_(steam) L_(v)+m_(s) xxS_(w)(100^@-theta^@)`
`=m_(w)L_(f) +m_(w)S_(w) (theta^@ - 0)`.
Promotional Banner

Topper's Solved these Questions

  • CALORIMETRY

    NARAYNA|Exercise Level- II (H.W)|11 Videos

  • CALORIMETRY

    NARAYNA|Exercise Level-V (Multi correct)|1 Videos

  • CIRCULAR MOTION

    NARAYNA|Exercise LEVEL II(H.W)|51 Videos

Similar Questions

Explore conceptually related problems

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 .

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 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)

How much steam at 100^(0)C is to be passed into water of mass 100g at 20^(0)C to raise its temperature by 5^(0)C ? (latent heat of steam is 540cal/gm )

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 .

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 )

How many calories are required to change one gram of 0^(@)C ice to 100^(@)C steam ? The latent heat of fusion is 80 cal // g and the latent heat of vaporization is 540 cal // g. The specific heat of water is 1.00 cal // (g K ) .

What is the amount of heat required (in calories) to convert 10 g of ice at -10^(@)C into steam at 100^(@)C ? Given that latent heat of vaporization of water is "540 cal g"^(-1) , latent heat of fusion of ice is "80 cal g"^(-1) , the specific heat capacity of water and ice are "1 cal g"^(-1).^(@)C^(-1) and "0.5 cal g"^(-1).^(@)C^(-1) respectively.