Thank you for visiting A 37 2 g sample of copper at 99 8 C is carefully placed into an insulated container containing 188 g of water at 18. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Answer:
Tâ‚‚ = 19.95°C
Explanation:
From the law of conservation of energy:
[tex]Heat\ Lost\ by\ Copper = Heat\ Gained\ by\ Water\\m_cC_c\Delta T_c = m_wC_w\Delta T_w[/tex]
where,
mc = mass of copper = 37.2 g
Cc = specific heat of copper = 0.385 J/g.°C
mw = mass of water = 188 g
Cw = specific heat of water = 4.184 J/g.°C
ΔTc = Change in temperature of copper = 99.8°C - Tâ‚‚
ΔTw = Change in temperature of water = Tâ‚‚ - 18.5°C
Tâ‚‚ = Final Temperature at Equilibrium = ?
Therefore,
[tex](37.2\ g)(0.385\ J/g.^oC)(99.8\ ^oC-T_2)=(188\ g)(4.184\ J/g.^oC)(T_2-18.5\ ^oC)\\99.8\ ^oC-T_2 = \frac{(188\ g)(4.184\ J/g.^oC)}{(37.2\ g)(0.385\ J/g.^oC)}(T_2-18.5\ ^oC)\\\\99.8\ ^oC-T_2 = (54.92) (T_2-18.5\ ^oC)\\54.92T_2+T_2 = 99.8\ ^oC + 1016.02\ ^oC\\\\T_2 = \frac{1115.82\ ^oC}{55.92}[/tex]
Tâ‚‚ = 19.95°C
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Rewritten by : Jeany
