100 g of liquid nitrogen at its boiling point of 77 k is stirred into a beaker containing 500 g of 15°c water.

if the nitrogen leaves the solution as soon as it turns to gas, how much water freezes?

the heat of vaporization of nitrogen is 48 cal/g and that of water is 80 cal/g.

a. noneb. 29 gc. 68 gd. 109 g

(a). none

Explanation:

Heat gained by nitrogen = Heat lost by water

Q₁ = Q₂ + Q₃ Equation 1

Where Q₁ = heat required for nitrogen to vaporize, Q₂ = heat required for to change temperature from 15 °C to 0 °C, Q₃ = heat required for water to vaporize.

Q₁ = m₁l₁ Equation 2

Q₂ =cm₂(T₁-T₂) Equation 3

Q₃ = m₃l₂ Equation 4

Note: For the water to freeze, Q₁ > Q₂

Where m₁ = mass of nitrogen, m₂ = mass of water in liquid state, m₃ = mass of water that freezes, l₁ = heat of vaporization of nitrogen, l₂ =  heat of vaporization of water, c = specific heat capacity of water, T₁ = initial temperature of water, T₂ = Final Temperature of water( Temperature at freezing point)

Given: m₁ = 100 g, m₂ = 500 g, T₁ = 15 °C, T₂ = 0 °C, l₁ = 48 cal/g, 80 cal/g

Constant: c = 1.00 cal/g.k

Substituting these values into equation 2, 3 and equation 4

Q₁ = 100 × 48 = 4800 Cal.

Q₂ = 1 × 500 × (15-0)

Q₂ = 7500 cal

Q₃ = m₃(80) = 80m₃ Cal

Substituting the values of Q₁, Q₂ and Q₃ into Equation 1

4800 = 7500 + 80m₃

Collecting like terms

80m₃ = 4800 - 7500

m₃ = -2700/80

m₃ = no ice formed, Because the mass of a substance can not be negative.

Therefore, The nitrogen evaporates but does not cause any change in temperature.

The right option is (a) none

c. newton's third law of motion