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klio [65]
3 years ago
6

You just measured a block of wood and obtained the following information: mass = 55.120 g length = 8.5 cm height = 4.3 cm width

= 3.3 cm Determine the volume and density of the wood block.
Physics
1 answer:
Misha Larkins [42]3 years ago
7 0

Answer:

Volume of the wood = 120.615 cm3

Density of the wood = 0.457 g/cm3

Explanation:

By definition, density is the mass of a substance per unit volume and its units are kg/m3, g/l etc.

Volume of a cuboid = length * breadth * height

Length = 8.5 cm

Breadth = 3.3 cm

Height = 4.3 cm

Volume of the wood = 8.5 * 3.3 * 4.3

= 120.615 cm3

Mass of the wood = 55.120 g

Density = mass/volume

= 55.120/120.615

= 0.457 g/cm3

= 457 kg/m3

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A 2.64-kg copper part, initially at 400 K, is plunged into a tank containing 4 kg of liquid water, initially at 300 K. The coppe
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Answer:

a) T_f=305.7049\ K

b) \Delta S=313.51\ J.K^{-1}

Explanation:

Given:

  • mass of copper, m_c=2.64\ kg
  • initial temperature of copper, T_{ic}=400\ K
  • specific heat capacity of copper, c_c=385\ J.kg^{-1}.K^{-1}
  • mass of water, m_w=4\ kg
  • initial temperature of water, T_{iw}=300\ K
  • specific heat capacity of water, c_w=4200\ J.kg^{-1}.K^{-1}

a)

<u>∵No heat is lost in the environment and the heat is transferred only between the two bodies:</u>

Heat rejected by the copper = heat absorbed by the water

2.64\times 385\times (400-T_f)= 4\times 4200\times (T_f-300)

T_f=305.7049\ K

b)

<u>Now the amount of heat transfer:</u>

Q=m_c.c_c.(T_{ic}-T_{f})

Q=2.64\times 385\times (400-305.7049)

Q=95841.5841\ J

∴Entropy change

\Delta S=\frac{dQ}{T}

\Delta S=\frac{95841.5841}{305.7049}

\Delta S=313.51\ J.K^{-1}

5 0
4 years ago
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