When 5.000 g of nahco3 is added to 600.0 g of water in a ‘coffee cup calorimeter’, the temperature decreases from 22.35°c to 18.94°c. a) calculate qcalorimeter, in joules, assuming the specific heat capacity of the solution is the same as pure water: 4.186 jg-1k-1. report your value to 4 significant figures, and do not include units in your answer.

Answer is: qcalorimeter is 8635.9273 J.

ΔT= 22.35°C - 18.94°C.

ΔT = 3.41°C = 3.41 K; temperature change.  

m(H₂O) = 600.0 g; mass of water.  

mr = 605.0; mass of solution.

cp= 4.186 J/g·K, specific heat capacity of the solution.  

qcal = mr · ΔT· cp.  

qcal = 605 g · 3.41 K · 4.186 J/g·K.  

qcal = 8635.9273 J; amount of heat absorbed.

the answers would be

the first one and the second one

(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