Correct Answer: Option A

Explanation

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)

  Combining Q1 and Q2

  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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