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Answer :
Answer:
To solve this problem, we need to apply the principle of conservation of energy. The heat lost by the lead soldier and the aluminium pedestal will be equal to the heat gained by the benzene as the system reaches thermal equilibrium at 15°C.
Step 1: Define the variables
Mass of lead (mₗ) = 500 g = 0.5 kg
Mass of benzene (mᵦ) = 500 g = 0.5 kg
Mass of aluminium pedestal (mₐ) = ?
Specific heat capacity of lead (cₗ) = 130 J/kg°C
Specific heat capacity of benzene (cᵦ) = 1716 J/kg°C
Specific heat capacity of aluminium (cₐ) = 909 J/kg°C
Initial temperature of lead and aluminium (Tᵢ) = 0°C
Initial temperature of benzene (Tᵦᵢ) = 22°C
Final temperature (T_f) = 15°C
Step 2: Write the heat transfer equations
Heat lost by lead and aluminium: The total heat lost (Qₗ + Qₐ) can be expressed as:
Q
lost
=m
l
⋅c
l
⋅(T
f
−T
i
)+m
a
⋅c
a
⋅(T
f
−T
i
)
Substituting the known values gives:
Q
lost
=0.5⋅130⋅(15−0)+m
a
⋅909⋅(15−0)
Q
lost
=0.5⋅130⋅15+m
a
⋅909⋅15
Q
lost
=975+13,635m
a
Heat gained by benzene: The heat gained by benzene (Qᵦ) is given by:
Q
gained
=m
β
⋅c
β
⋅(T
f
−T
βi
)
Substituting the known values gives:
Q
gained
=0.5⋅1716⋅(15−22)
Q
gained
=0.5⋅1716⋅(−7)=−6006
Step 3: Set the heat lost equal to the heat gained
According to the conservation of energy:
Q
lost
=−Q
gained
Thus, we have:
975+13,635m
a
=6006
Step 4: Solve for the mass of the aluminium pedestal (mₐ)
Now we can rearrange the equation:
13,635m
a
=6006−975
13,635m
a
=5028
m
a
=
13,635
5028
≈0.368kg
Step 5: Conclusion
The mass of the aluminium pedestal is approximately 0.368 kg or 368 g.
This calculation demonstrates the balance of heat transfer in the system, showing how energy is conserved as the toy soldier and pedestal heat up while the benzene cools down.
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