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Question: What is the rise in temperature of a collective drop when initially 1gm and 2gm drops travel with ve...

What is the rise in temperature of a collective drop when initially 1gm and 2gm drops travel with velocities 10cm/sec  10cm/sec\; and 15cm/sec15cm/sec
A) 6.6×10−3 ∘C6.6 \times {10^{ - 3}}\,^\circ C
B) 66×10−3 ∘C66 \times {10^{ - 3}}\,^\circ C
C) 660×10−3 ∘C660 \times {10^{ - 3}}\,^\circ C
D) 6.6 ∘C6.6\,^\circ C

Explanation

Solution

In this solution, we will calculate the kinetic energy of the drops before and after the collision. The decrease in kinetic energy will contribute to raising the temperature of the collective drop.

Formula used:
Q=msΔTQ = ms\Delta T where QQ is the energy needed to heat up a liquid by ΔT\Delta T temperature difference and ss is the specific heat capacity of the liquid

Complete step by step answer:
We’ve been given the temperature of the drops as 10cm/sec  =0.1 m/s10cm/sec\; = 0.1\,m/s and 15 cm/sec  =0.15 m/s15\,cm/sec\; = 0.15\,m/s. Let us start by calculating the velocity of the new drop. When the drops combine, the law of conservation of momentum holds true since there is no external force acting on the system. So, applying the law of conservation of momentum before and after coagulation, we get
m1v1+m2v2=m′v′{m_1}{v_1} + {m_2}{v_2} = m'v'
Substituting m1=1gm=1×10−3 kg{m_1} = 1gm = 1 \times {10^{ - 3}}\,kg, m2=2gm=2×10−3 kg{m_2} = 2gm = 2 \times {10^{ - 3}}\,kg, v1=0.1 m/sec  {v_1} = 0.1\,m/sec\; and v2=0.15 m/sec{v_2} = 0.15\,m/sec, and m′=m1+m2=3×10−3 kgm' = {m_1} + {m_2} = 3 \times {10^{ - 3}}\,kg we get
(1×0.1)+(2×0.15)=(1+2)×v′\left( {1 \times 0.1} \right) + (2 \times 0.15) = (1 + 2) \times v'
Which gives us
v=403 ×10−2 m/sv = \dfrac{{40}}{3}\, \times {10^{ - 2}}\,m/s
Now the kinetic energy of the two drops before the coagulation will be
Kinit=12×1×10−3×(0.1)2+12×2×10−3×(0.15)2{K_{init}} = \dfrac{1}{2} \times 1 \times {10^{ - 3}} \times {\left( {0.1} \right)^2} + \dfrac{1}{2} \times 2 \times {10^{ - 3}} \times {\left( {0.15} \right)^2}
⇒Kinit=275×10−7 J\Rightarrow {K_{init}} = 275 \times {10^{ - 7}}\,J
Similarly, the kinetic energy after coagulating will be
Kfinal=123×10−3×(403 ×10−2 )2{K_{final}} = \dfrac{1}{2}3 \times {10^{ - 3}} \times {\left( {\dfrac{{40}}{3}\, \times {{10}^{ - 2}}\,} \right)^2}
⇒Kfinal=8003×10−7 J\Rightarrow {K_{final}} = \dfrac{{800}}{3} \times {10^{ - 7}}\,J
This difference in kinetic energy will raise the temperature of the water. The difference in kinetic energies will be
ΔK=(275−8003)×10−7\Delta K = \left( {275 - \dfrac{{800}}{3}} \right) \times {10^{ - 7}}
⇒ΔK=253×10−7 J\Rightarrow \Delta K = \dfrac{{25}}{3} \times {10^{ - 7}}\,J
This difference will heat the drop according to the equation of specific heat as
ΔK=msΔT\Delta K = ms\Delta T
So, substituting ΔK=253×10−7 J\Delta K = \dfrac{{25}}{3} \times {10^{ - 7}}\,J, m=3×10−3 kgm = 3 \times {10^{ - 3}}\,kg and s=4.2s = 4.2 for water, we get
253×10−7 =3×10−3 ×4.2×ΔT\dfrac{{25}}{3} \times {10^{ - 7}}\, = 3 \times {10^{ - 3}}\, \times 4.2 \times \Delta T
So, the change in temperature will be
ΔT=259×4.2=66×10−3 ∘C\Delta T = \dfrac{{25}}{{9 \times 4.2}} = 66 \times {10^{ - 3}}\,^\circ C

Hence the correct choice is option (B).

Note: Here we have assumed that all the difference in the kinetic energy will go to increasing the temperature of the drop however this is an ideal gas. In reality, there is some energy used when combining and as a result full conversion of kinetic to heat energy doesn’t occur.