Thank you for visiting A 28 4 g sample of an unknown metal is heated to 39 4 C then placed in a calorimeter containing 50 0 g of. 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 :
The specific heat of the unknown metal in the problem is calculated to be 0.898 J/g°C. This corresponds to option B. The calculation follows the principle of conservation of energy.
To solve this, we use the principle of conservation of energy which states that the heat lost by the metal will be equal to the heat gained by the water:
Heat gained by water, Q = mass of water * specific heat of water * change in temperature.
Q = 50.0 g * 4.18 J/g°C * (23.00°C - 21.00°C)
Q = 50.0 g * 4.18 J/g°C * 2.00°C
Q = 418 J
Next, we calculate the specific heat of the metal (cmetal) using the formula:
Q = m * cmetal * ΔT
418 J = 28.4 g * cmetal * (23.00°C - 39.4°C)
418 J = 28.4 g * cmetal * (-16.4°C)
cmetal = 418 J / (28.4 g * -16.4°C)
cmetal ≈ 0.898 J/g°C
Therefore, the specific heat of the metal in this problem is 0.898 J/g°C, which corresponds to option B.
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Rewritten by : Jeany
