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A metal piece of 50 g specific heat 0.6 cal/gºC initially at 120ºC is dropped in 1.6 kg of water at 25ºC. Find the final temperature or mixture. |
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Answer» m1c1 (θ1 – θ) = m2c2 (θ – θ2) ∴ c2 = 1 cal/gmºC ∴ 50 × 0.6 × (120 – θ) = 1.6 × 103 × 1 × (θ – 25) θ = 26.8ºC |
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