Question

What is the rise in tempreture of a collective drop when initially 1gm and 2 gm drops travel with velocities 10 and 15?

Answer 1

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