Question on: JAMB Physics - 2018
Water of mass 150g at 60\(^o\)c is added to 300g of water at 20\(^o\)c and the mixture is well stirred. Calculate the temperature of the mixture.(neglect heat losses to the surroundings)
A
33\(^o\)c
B
40\(^o\)c
C
25\(^o\)c
D
10\(^o\)c
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Correct Option: A
Water 1 => M = 150g = 0.15kg,
 θ\(_2\) = 60\(^o\)C; θ\(_1\) = θ
 Q\(_1\) = MC (θ\(_2\) – θ\(_1\))
 = 0.15C (60 - θ)
 Water 2 => M = 300g = 0.3kg
 θ1 = 20\(^o\)C, θ\(_2\) = θ
 Q2 = MC (θ\(_2\) – θ\(_1\))
 = 0.3C (θ – 20)
 Combine Q1 & Q2, we get:
 0.15C (60 - θ) = 0.3C (θ – 20)
 9 – 0.15θ = 0.3θ – 6
 0.3θ + 0.15θ = 9 + 6
 0.45 θ = 15
 θ = 33.33\(^o\)C
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