Question

In a insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Find the final temperature of the mixture (in kelvin.) Given, L_(fusion) = 80 cal//g = 336 J//g L_("vaporization") = 540 cal//g = 2268 J//g s_(ice) = 2100 J//kg-K = 0.5 cal//g-K and s_("water") = 4200 J//kg-K = 1cal//g-K .

Answer 1

answer : final temperature of the mixture is 273K.

given, mass of steam, m₁ = 0.05 kg = 50g at 100°C

mass of ice, m₂ = 0.45 kg = 450g at -20°C

lost heat by steam to change into water, H₁ = m₁Lv

= 50g × 2268 J/g

= 113400 J

heat lost by water to decrease temperature, H₂ = m₁s₁∆T

= 50g × 4.2 J/g × (100°C - 0°C)

= 50 × 4.2 × 100

= 21000 J

gained heat by ice , Q = m₂s₂∆T

= 450g × 2.1 J/g × (0 + 20°C)

= 450 × 2.1 × 20 J

= 18900 J

heat gained by ice to change ice into water, Q' = m₂Lf

= 450g × 336 J/g

= 151200 J

The temperature will be H₁ + H₂ < Q + Q'

so the final temperature of the mixture is 273K.

Answer 2

Explanation:

HERE IS UR SOLUTION.

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