When a 120 g sample of aluminum absorbs 9612 of heat energy, its temperature increases from 25°C to 115°C. Find the specific hea t of aluminum
1 answer:
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Answer: </h3>
0.89 J/g°C
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Concept tested: Quantity of heat
We are given;
Mass of the aluminium sample is 120 g Quantity of heat absorbed by aluminium sample is 9612 g Change in temperature, ΔT = 115°C - 25°C = 90°C
We are required to calculate the specific heat capacity;
We need to know that the quantity of heat absorbed is calculated by the product of mass, specific heat capacity and change in temperature. That is;
Q = m × c × ΔT
Therefore, rearranging the formula we can calculate the specific heat capacity of Aluminium. Specific heat capacity, c = Q ÷ mΔT
= 9612 J ÷ (120 g × 90°C)
= 0.89 J/g°C
Therefore, the specific heat capacity of Aluminium is 0.89 J/g°C
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V₂ = 2.91 L
Explanation:
Given data:
Initial volume = 3.50 L
Initial temperature = 90.0°C (90+273 = 363 K)
Final temperature = 30.0 °C ( 30 +273 = 303 K)
Final volume = ?
Solution:
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
V₁/T₁ = V₂/T₂
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V₂ = 2.91 L
