Question

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

Answer 1

Answer:

Option A: none

Explanation:

Specific heat capacity is the amount of heat energy required to raise the temperature of a substance per unit of mass.

From the question, the parameters given are; The heat of vaporization of nitrogen= 48 cal/g and that of water is 80 cal/g.

Using the formulae; Specific heat capacity,c= Q/ m× ∆T,----------–-----------------------------------------------------

STEP ONE: We will have to calculate all the energy numbers; 77k is approximately the boiling point of Nitrogen.

Energy required to decrease water from 15°C to 0°C = E(1).

0.1×10^3 g× 48 cal/gram..

= 4800 cal

Energy require to vaporize Nitrogen=E(2).

= 80 cal per gram×0.15(15-0)

= 180 cal

Energy required to decrease water from 15°C to 0°C is higher than that of the energy to vaporize Nitrogen, N2.

STEP 2: 180/4800× 15 = 0.5625

Therefore; 15-0.5625 =14.43

Answer= 14.43gC

Answer 2

Answer:

(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