Correct Question :
Mass of water = 50.003g
Temperature of water= 24.95C
Specific heat capacity for water = 4.184J/g C
Mass of metal = 63.546 g
Temperature of metal 99.95°C
Specific heat capacity for metal ?
Final temperature = 32.80°C
In an experiment to determine the specific heat of a metal student transferred a sample of the metal that was heated in boiling water into room temperature water in an insulated cup. The student recorded the temperature of the water after thermal equilibrium was reached. The data we shown in the table above. Based on the data, what is the calculated heat absorbed by the water reported with the appropriate number of significant figures?
Answer:
1642 J
Explanation:
Given:
Mass of water = 50.003g
Temperature of water= 24.95C
Specific heat capacity for water = 4.184J/g C
Mass of metal = 63.546 g
Temperature of metal 99.95°C
Specific heat capacity for metal ?
Final temperature = 32.80° C
To calculate the heat absorbed by water, Q, let's use the formula :
Q = ∆T * mass of water * specific heat
Where ∆T = 32.80°C - 24.95°C = 7.85°C
Therefore,
Q= 7.85 * 50.003 * 4.184
Q = 1642.32 J
≈ 1642 J
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