Question

What is rise in temperature of a collective drop when initially 1 gm and 2 gm drop travel with velocities 10 cm/s and 15 cm/s?

Answer 1

answer : 6.6 × 10^-5 °C

explanation : we have to find temperature of a collective drop when initially 1gm and 2gm drops travel with velocities 10cm/s and 15cm/s respectively.

applying law of conservation of linear momentum,

before coagulation = after coagulation

1 × 10 + 2 × 15 = (1 + 2) × u

u = 40/3 cm/s

kinetic energy before coagulation of drops = 1/2 × 1 × (10)² + 1/2 × 2 × (15)²

= 50 + 225 = 275 erg

[we know, 1g.cm² = 1erg ]

kinetic energy after coagulation of drops = 1/2 (1 + 2) × (40/3)²

= 1/2 × 3 × 1600/9

= 800/3 erg

so, energy converted into heat = kinetic energy before coagulation of drops - kinetic energy after coagulation of drops

= 275 - 800/3 = 25/3 erg

= 25/3 × 10^-7 J [ we know, 1 erg = 10^-7 J]

we know, heat = msT

where , m is mass , s is specific heat capacity and T is temperature.

S = 4.2 J/Kg/°C , m = 3g = 3 × 10^-3 kg

so, 25/3 × 10^-7 = 3 × 10^-3 × 4.2 × T

or, T = 25/(9 × 4.2) = 6.6 × 10^-5 °C

Answer 2

Given mass of drop 1 m₁ = 1 gm = 1×10⁻³ kg

mass of drop 2 m₂ = 2 gm = 2×10⁻³ kg

Velocity of drop 1 v₁ = 10 cm/s = 10× 10⁻² m/s  

Velocity of drop 2 v₂ = 15 cm/s = 15× 10⁻² m/s

Now conservation of momentum we get,

m₁v₁+m₂v₂ = (m₁+m₂) V

V = (m₁v₁+m₂v₂)/ (m₁+m₂)

V =(1×10⁻³×10×10⁻² + 2×10⁻³×15×10⁻²)/(1×10⁻³ + 2×10⁻³)

V = (40/3)×10⁻² m/s

From work energy theorem we have

ΔK.E = Q(Heat) --------(A)

Also we know Q = m c ΔT  -------(B)

ΔK.E = K.E₂ -K.E₁

= {1/2 (0.001+0.002)(40/3×10⁻²)²} - {1/2 (0.002+0.002)(40/3×10⁻²)²}

= 88.81 × 10⁻⁷ J

Reducing the above found value in B we get,

(88.81 × 10⁻⁷)/(0.003× 4.2) = ΔT

ΔT = 7.05 ⁰ C