How much energy will be released when 600g of steam at 100⁰C is condensed to water at 100⁰C ?

Answer

53.6 KJ

Explanation:

The energy released came from two discrete processes: the steam condenses to release some latent heat of condensation at  100 l o C the water cools down to  0 l o C  to  100 l o C  without solidfying. The amount of energy released in the first process is dependent on the  L v  latent heat of vaporization for water and the mass of the sample: E ( phase change ) = m ⋅ L v = 20 l g × 2 , 260 l J ⋅ g − 1  E ( phase change ) = 45 , 200 l J On the other hand, the amount of energy released in the second process depends on both the specific heat of water, the mass of the sample, and the magnitude of the change in temperature. E ( cooling ) = m ⋅ c ⋅ Δ T  E ( cooling ) = 20 l g ⋅ 4.2 l J ⋅ g − 1 ⋅ K − 1 ⋅ ( 100 l K − 0 l K )  E ( cooling ) = 8 , 400 l J  Taking the sum of the energy change of the two processes gives the total amount of energy released: E ( released ) = E ( phase change ) + E ( cooling ) = 53 , 600 l J = 53.6 l kJ

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied