In a container of negligible heat capacity, 200 gm ice at 0°C and 100 gm steam at 100°C are added to 200 gm of water that has temperature 55°C. Assume no heat is lost to the surroundings and the pressure in the container is constant 1.0 atm. (Latent heat of fusion of ice = 80 cal/gm, Latent heat of vaporization of water = 540 cal/gm, Specific heat capacity of ice = 0.5 cal/gm-K, Specific heat capacity of water = 1 cal/gm-K)
(i) What is the final temperature of the system?
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) : As steam has comparatively large amount of heat to provide in the form of latent heat we
check what amount of heat is required by the water and ice to go up to 100°C, that is
(ii) : (m i L + m i S w Δ T) + m w . S w . Δ T
= [(200 × 80) + (200 × 1 × 100)] + (200 × 1× 45)
= 45,000 cal.
That is given by m mass of steam, then
m s .L = 45,000
m s = 
= 
= 83.3 gm
therefore 83.3 gm steam converts into water of 100°C.
Total water = 200 + 200 + 83.3 = 483.3 gm
(iii) : P steam left = 16.7 gm.
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