Calculate the heat lost by the hot water using your results for experiment 3: measuring the enthalpy of fusion of water. heat must be expressed in kilojoules (kj). q= (4.18 j/g °c) x (1 kj/1000 j) x mass of water (g) x δt

7.399 kJ

Explanation:

The heat lost (Q) by the hot water is the product of the mass (m) of hot water, temperature difference (ΔT) and the (c) specific heat capacity of water. Mathematically,

Q = m*c*ΔT

The final volume of hot water = 30 ml, and the density of water = 1 g/ml

Thus, mass (m) = 30*1 = 30g

ΔT = final temperature - initial temperature = 60 - 1 = 59 °C

Specific heat capacity of water = 4.18 J/g °C

Therefore

Q = m*c*ΔT = 4.18*30*59 = 7398.6 J

To convert from J to kJ, we will divide by 1000:

Q = 7398.6/1000 = 7.399 kJ

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(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