Thank you for visiting A 51 02 g piece of unknown metal at 99 8 degrees Celsius was placed in a calorimeter with 75 mL of water at 21. 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 :
Specific heat capacity of the unknown metal = 0.551 J/g°C.
In this problem, the specific heat of the unknown metal is to be found. The metal is transferred to a calorimeter that contains 75 mL of water at 21.0 degrees Celsius, and the final temperature of the system is recorded at 24.0 degrees Celsius. Using the formula of specific heat capacity, q = mcΔT, we can write the equation, Heat lost by the metal = Heat gained by the water m × c × ΔT = m × c × ΔT Where, m = mass of the metal c = specific heat capacity of the metalΔT = change in temperature = (Final temperature of the system) - (Initial temperature of the system)Substituting the given values, m = 51.02 gc = ΔT = 24.0 - 21.0 = 3.0 °C (since the temperature increased)The mass of the water can be determined using its density, d = 1 g/mL.
Therefore, m (water) = Volume × density = 75 mL × 1 g/mL = 75 g The specific heat of water is 4.18 J/g°C. We have, m (water) = 75 gc (water) = 4.18 J/g°CΔT = 24.0 - 21.0 = 3.0 °C We have calculated the heat lost by the metal and the heat gained by the water to be the same. Therefore, the formula can be equated as follows: m × c × ΔT = m (water) × c (water) × ΔT (water)51.02 × c × 3.0 = 75 × 4.18 × 3.0c = (75 × 4.18 × 3.0) / (51.02 × 3.0)c = 0.551 J/g°C.
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