Water of mass 150g at 60 o c is added to 300g o... - JAMB Physics 2018 Question
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
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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