How much heat is absorbed by 15.5g of steam to turn into liquid water when its temperature is decreased from 120.0°c to 100.0°c?

- 618.8 J.

Explanation:

Q = m.c.ΔT,

where, Q is the amount of heat released by steam (Q = ??? J).

m is the mass of the steam (m = 15.5 g).

c is the specific heat of the steam (c of steam = 1.996 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 120.0°C = - 20.0°C).

∴ Q = m.c.ΔT = (15.5 g)(1.996 J/g.°C)(- 20.0°C) = - 618.8 J.

(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