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a 1.00kg piece of aluminum metal at 90 degrees Celsiusis placed in 4 L (=4.00kg)of water at 25°C determine the final temperature



Answer :

From the basic "heat lost by hot object=heat gained by colder object" principle, we have
m1c1ΔT1=m2c2ΔT2
where m1= 1kg
m2=4kg
c1=900J/kg k
c2=4200J/kg k 
With this information at hand we have
m1c1(90-T)=m2c2(T-25)
after substituting the given values we can find that
T=28.3^{0}c
Lanuel

By applying the Law of Conservation of Energy, the final temperature of water is equal to 28.63°C

Given the following data:

  • Mass of aluminum metal = 1.00 kg
  • Volume of water = 4 L
  • Mass of water = 4.00 kg
  • Temperature of aluminum metal = 90.0°C
  • Initial temperature of water = 25.0°C

Scientific data:

  • Specific heat capacity of water = 4200 J/kg°C
  • Specific heat capacity of aluminum = 900 J/kg°C

Mathematically, quantity of heat energy is given by the formula;

[tex]Q=mc\theta[/tex]

Where:

  • Q represents the quantity of heat energy.
  • m represents the mass of an object.
  • c represents the specific heat capacity.
  • ∅ represents the change in temperature.

According to the Law of Conservation of Energy, we have:

The quantity of heat energy lost by the aluminum metal = The quantity of heat energy gained by the water.

[tex]Q_{lost} = Q_{gained}\\\\m_ac_a\theta_a = m_wc_w\theta_w\\\\1(900)(90 - T_f) = 4(4200)(T_f - 25)\\\\81000-900T_f =16800T_f - 420000\\\\16800T_f+900T_f=420000 + 81000\\\\T_f=\frac{501000}{17500} \\\\T_f=28.63[/tex]

Final temperature = 28.63°C

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