Scoring Scheme: 3-3-2-1 Part II. For each trial, enter the amount of heat lost b

Question

Scoring Scheme: 3-3-2-1 Part II. For each trial, enter the amount of heat lost by the metal, qmetal- Hint: The specific heat of water is 4 184 JIg°C. Be careful of your algebraic sign here and remember that the change in temperature is equal to une final temperature minus the initial temperature. Your answer should be reported to 4 digits. Note: You should always carry 1 or 2 extra digits beyond the number of significant figures until your final calculation Trial #MassMetal Timetal Masscool water T 70.200 70.500 70.300 ter 22.5 23.0 23.2 qmetal #2: #3 183.100 183.100 183.100 103.0 101.0 102.0 36.5 37.5 36.5 Submit Answer/Continue Quit Post-Lab

Explanation / Answer

Heat gained by water= mass of water (heated)*Specific heat of water*change of temperature=Mass(metal)*Cs*(Tf-Tiwater)

trial 1)

Heat lost by metal=Heat gained by water( law of conservation of energy)

Heat lost=(70.200g)*(4.184J/g C)*(36.5-22.5)cdeg C=4112.0352 J=4112.0J

qmetal=4112.0 Joules

trial 2)

Heat lost by metal=Heat gained by water( law of conservation of energy)

Heat lost=(70.500g)*(4.184J/g C)*(37.5-23.0) deg C=4112.0352 J=4112.0J

qmetal=4277.094J=4277.1J

trial 3)

Heat lost by metal=Heat gained by water( law of conservation of energy)

Heat lost=(70.300g)*(4.184J/g C)*(36.5-23.2) deg C=4112.0352 J=4112.0J

qmetal=3911.99816 J=3912.0 J

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