0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

An Insulated vessel contains 0.4 kg of water at 0. A piece of 0.1 kg ice at 15C is put into it and steam at 100C is bubbled into it until all ice is melted and finally the contents are liquid water at 40C. Assume that the vessel does not give or take any heat and there is no loss of matter and heat to the surroundings. Specific heat of ice is 2.2×103Jkg1K1. heat of fusion of water is 333×103Jkg1 heat of vaporization of water is 2260×103Jkg1 The amount of steam that was bubbled into the water is :
  1. 34.7 gram
  2. 236 gram
  3. 0.023 gram
  4. 48.01 gram

A
236 gram
B
34.7 gram
C
0.023 gram
D
48.01 gram
Solution
Verified by Toppr

Heat gained by water of 0.4kg to change its temperature=q1
q1=mcΔT
Where,
m=mass
C= Specific capacity
ΔT= Temperature difference
q1=0.4×103×40×1
=16000 calories (specific heat of water= 1cal/gm°C)
Heat gained by0.4kg of ice to change its temperature =q2
q2=mcΔT (Specific capacity of ice=12cal/gm°C)
=0.1×12×15×1000
=0.75calories
=750calories
Heat gained by .01kg of ice to convert into water=mL
q3=mL (L=Latent heat of fusion)
=0.1×80×1000
= 8000calories
Heat gained by 0.14kg water to change temperature from 0°C to 40°C=mcΔT=q4
q4=0.110×1000×1×40
=4000calories
Total heat gained =q1+q2+q3+q4
=16000+750+8000+4000
=28750calories
Heat lost by steam to change state to liquid=q5
q5=mL (Latent heat of vaporization)
q5=m×540calories
Heat lost by water to change temperature from 100°C to 40°C= q6=mcΔT
=m×1×60
=60m calories
At equilibrium heat gained = heat lost
28750=600m
m=28750600gm
m=48.01gm



Was this answer helpful?
0
Similar Questions
Q1
An Insulated vessel contains 0.4 kg of water at 0. A piece of 0.1 kg ice at 15C is put into it and steam at 100C is bubbled into it until all ice is melted and finally the contents are liquid water at 40C. Assume that the vessel does not give or take any heat and there is no loss of matter and heat to the surroundings. Specific heat of ice is 2.2×103Jkg1K1. heat of fusion of water is 333×103Jkg1 heat of vaporization of water is 2260×103Jkg1 The amount of steam that was bubbled into the water is :
View Solution
Q2
A copper calorimeter of mass 100gm contains 200gm of a mixture of ice and water. Steam at 100C under normal pressure is passed into the calorimeter and the temperature of the mixture is allowed to rise to 50C. If the mass of the calorimeter and its contents is now 330gm, what was the ratio of ice and water in the beginning? Neglect heat losses.
Given: Specific heat capacity of copper =0.42×103Jkg1K1 ,
Specfic heat capacity of water =4.2×103Jkg1K1
Specific heat of fusion of ice =3.36×105Jkg1
Latent heat of condensation of steam =22.5×105Jkg1
View Solution
Q3
A piece of ice of mass 40 g at 0oC is added to 200 g of water at 50C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water = 4200Jkg1K1 and specific latent heat of fusion of ice = 336×103Jkg1 :
View Solution
Q4
A Carnot refrigerator absorbs heat from water at 0 and gives it to a room at 27. When it converts 2kg of water at 0 into ice at 0, the work done is (Latent heat of fusions of ice =333×103Jkg1
View Solution
Q5
1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 103 J kg−1 and latent heat of vaporization of water = 2.26 × 106 J kg−1.
View Solution