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Thepotemich [5.8K]
2 years ago
8

Bumper car A (281 kg) moving

Physics
1 answer:
solmaris [256]2 years ago
4 0

use consevation of linear momentum

  • m1v1+m2v2=(M1+M2)V3
  • 281(2.82)+209(-1.72)=(209+281)V2
  • 792.42-359.48=490v3
  • 432.9=490v3
  • V3=0.88m/s
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A 0.350 kg block at -27.5°C is added to 0.217 kg of water at 25.0°C. They come to equilibrium at 16.4°C. What is the specific he
Gekata [30.6K]

The specific heat capacity of the block is 508J/kg^{\circ}C

Explanation:

As the block is placed into the water, heat energy is transferred from the water (which is at higher temperature) to the block (which is at lower temperature), until the block and the water are in thermal equilibrium (= same temperature).

Therefore, we can write:

Q_{water}=Q_{block}

Where

Q_{water}=m_w C_w (T_w-T_{eq}) is the heat energy released by the water, where

m_w = 0.217 kg is the mass of the water

C_w = 4186 J/kg^{\circ}C is the water heat specific capacity

T_w = 25.0^{\circ} is the initial temperature of the water

T_{eq}=16.4^{\circ} is the temperature at equilibrium

Substituting,

Q_{water}=(0.217)(4186)(25.0-16.4)=7812 J

Now we can write the heat energy absorbed by the block as

Q_{block}=m_b C_b(T_{eq}-T_b)

where

m_b=0.350 kg is the mass of the block

C_b is the specific heat capacity of the block

T_b = -27.5^{\circ} is the initial temperature of the block

And solving for C_b,

C_b=\frac{Q_{block}}{m_b(T_{eq}-T_b)}=\frac{7812}{(0.350)(16.4-(-27.5))}=508J/kg^{\circ}C

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

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