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onyebuchinnaji

Answer:

The fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

Explanation:

Given;

initial temperature of the liquid, t₁ = 76.3  +/-  0.4⁰C

final temperature of the liquid, t₂ = 67.7  +/-  0.3⁰C

The change in temperature of the liquid is calculated as;

Δt = t₂  -  t₁

Δt = (67.7 - 76.3)  +/-  (0.3 - 0.4)

Δt = (-8.6)  +/-  (-0.1)

Δt = 8.6 +/- 0.1 ⁰C

Therefore, the fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

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