A 135 g sample of H20 at 85°C is cooled. The water loses a total of 15 kJ of energy in the cooling process. What is the final temperature of the water? The specific heat of water is 4.184 J/g.°C.

A. 112°C

B. 58°C

C. 70°C

D. 84°C

E. 27°C

B. 58°C

Explanation:

Hello,

In this case, the relationship among heat, mass, specific heat and temperature for water is mathematically by:

In such a way, solving for the final temperature we obtain:

Therefore, we final temperature is computed as follows, considering that the involved heat is negative as it is lost for water:

Thereby, answer is B. 58°C .

Regards.

what's the question?

explanation:

(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