Question

HELP 80 points

Answer 1

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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