Question

20g of steam at 100 degree C is passed into 100g of ice at 0 degree C. Find the resultant temperature if specific latent heat of steam is 540 cal/g., specific latent heat of ice is 80 cal/g and specific heat of water is 1 cal/g℃

Answer 1

Ice at 0°C has a total latent heat of Q=m*L = 540*80 = 43200 cal. Now for ... So final temp of mixture is 0°c and mixture is saturated water.

Answer 2

Answer:m1=mass of steam

L1=latent heat of vaporization

S=specific heat of water

m2=mass of ice

L2=latent heat of fusion

T1= temperature of steam

T2=temperature of ice

T3=resultant temperature

m1L1+m1S(T1-T3)= m2L2+m2S(T3-T2)

m1= 20g

L1=540 cal/g°C

S=1 cal/g°C

m2=100g

L2=80 cal/g°C

T1=100°C

T2=0°C

20×540+20×1×(100-T3)=100×80+100×1×(T3-0)

10800+2000-20T3=8000+100T3

12800-8000=120T3

T3=4800/120

T3=40°C

Explanation: