Home
Class 12
CHEMISTRY
A mole of steam is condensed at 100^(@) ...

A mole of steam is condensed at `100^(@)` C, the water is cooled to `0^(@)`C and frozen to i.e . What is the difference in entropies of the stem and ice? The heat of vaporization and fusion are `540 cal" "gm^(-1)` and `80 cal" "gm ^(-1)` respectively . Use the average heat capacity of liquild water as `1 cal" "gm^(-1)` `degree^(-1)`.

A

Entropy change during the condensation of steam is `-26.06 cal//^(@)C`

B

Entropy change during cooling of water from `100^(@)C "to" 0^(@)C "is" - 5.62 cal//^(@)C`

C

Entropy change during freezing of water at `0^(@)C "is" -5.27 cal//^(@)C`

D

Total entropy changwe is `-36.95 cal//^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
Promotional Banner

Topper's Solved these Questions

  • THERMODYNAMICS

    AAKASH INSTITUTE|Exercise SECTION-D|9 Videos

  • THERMODYNAMICS

    AAKASH INSTITUTE|Exercise SECTION-E|10 Videos

  • THERMODYNAMICS

    AAKASH INSTITUTE|Exercise SECTION-B|20 Videos

  • THE SOLID STATE

    AAKASH INSTITUTE|Exercise Assignment (SECTION - D) (ASSERTION-REASON TYPE QUESTION)|20 Videos

Similar Questions

Explore conceptually related problems

One mole of ice is melted at 0^(@)C and then is heated to 100^(@)C . What is the difference in entropies of the steam and ice? The heats of vaporisation and fusion are 540 cal g^(-1) and 80 cal g^(-1) respectively. Use the average heat capacity of liquid water as 1cal g^(-1) degree^(-1)

If x gm of steam at 100^(@) C is mixed with 5x gm of ice at 0^(@)C , calculate the final temperature of resultiong mixture. The heat of vapourisation and fusion are 540 cal gm^(-1) and 80 cal gm^(-1) respectively.

How much heat is required to change 10g ice at 10^(@)C to steam at 100^(@)C ? Latent heat of fusion and vaporisation for H_(2)O are 80 cl g^(-1) and 540 cal g^(-1) , respectively. Specific heat of water is 1cal g^(-1) .

Steam at 100°C is passed into 100 g of ice at 0°C . Finally when 80g ice melts the mass of water present will be [Latent heat of fusion and vaporization are 80 cal g^(-1) and 540 cal g^(-1) respectively, specific heat of water = 1 cal g^(-1) °C^(-1) ]

How much heat is required to change 5 gram ice (0^@C) to steam at 100^@C ? Latent heat of fusion and vaporization for water are 80 cal/g and 540 cal/g respectively . Specific heat of water is 1 cal/ g/k.

A vessel contains a small amount of water at 0^(@)C . If the air in the vessel is rapidly pumped out, it causes freezing of the water. Why? What percentage of the water in the container can be frozen by this method? Latent heat of vaporization and fusion are L_(V) = 540 cal g^(-1) and L_(f) = 80 cal g^(-1) respectively.

100 gm of water at 20^@C is converted into ice at 0^@C by a refrigerator in 2 hours. What will be the quantity of heat removed per minute? Specific heat of water = 1 cal g^(-1) C^(-1) and latent heat of ice = 80 cal g^(-1)

Calculate the amount of heat required in calorie to change 1 g of ice at -10^(@)C to steam at 120^(@)C . The entire process is carried out at atmospheric pressure. Specific heat of ice and water are 0.5 cal g^(-1) .^(@)C^(-1) and 1.0 cal g^(-1) .^(@)C^(-1) respectively. Latent heat of fusion of ice and vaporization of water are 80 cal g^(-1) and 540 cal g^(-1) respectively. Assume steam to be an ideal gas with its molecules having 6 degrees of freedom. Gas constant R = 2 cal mol^(-1) K^(-1) .