A30.5 g sample of an alloy at 95.0°c is placed into 49.3 g water at 24.3°c in an insulated coffee cup. the heat capacity of the coffee cup (without the water) is 9.2 j/k. if the final temperature of the system is 31.1°c, what is the specific heat capacity of the alloy? (c of water is 4.184 j/g×k)

0.752 J/g*K

Explanation:

The heat lost by the alloy (which is negative) must be equal to the heat gained by the water and the coffee cup:

-Qa = Qw + Qc

-ma*ca*ΔTa = mw*cw*ΔTw + C*ΔTc

Where, m is the mass, c is the specific heat capacity, C is the heat capacity of the coffee cup, ΔT is the change in temperature, a represents the alloy, and w the water.

The coffee cup has initial temperature equal to the water, then:

-30.5*ca*(31.1 - 95.0) = 49.3*4.184*(31.1 - 24.3) + 9.2*(31.1 - 24.3)

1948.95ca = 1465.20

ca = 0.752 J/g*K

(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