243.92J
Explanation:
Given parameters:
Mass of the aluminum cup = 42.14g
Initial temperature of the cup, t₁ = 20°C
Final temperature of the cup t₂ = 26.41°C
Specific heat capacity of aluminium = 0.903 J g⁻¹ °C⁻¹
Unknown:
Change in heat of the aluminium cup, q = ?
Solution:
The heat lost during the experiment is the heat gained by the aluminum cup.
The quantity of heat is found using the expression below:
q = m C Δt
m is the mass of the cup
C is the specific heat of the cup
Δt is the change in heat
q = 42.14 x 0.903 x (26.41 - 20) = 42.14 x 0.903 x 6.41
q = 243.92J
Learn more:
Specific heat capacity brainly.com/question/7210400
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