What is rise in temperature of collective drop when initially 1 gram and 2gram drops travel with velocity 10 cm per second and 15 cm per second respectively
Answers
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