First run Second run Mass of metal A.2 Mass of cup A.3 Mass of cup + water A.4 T
ID: 717973 • Letter: F
Question
First run Second run Mass of metal A.2 Mass of cup A.3 Mass of cup + water A.4 Temperature of boiling water bath 52.00590 1.500"( 130 c 12.oC 12.0c 5.0 A.5 Initial temperature of water in cup A.6 Final temperature of water and metal in cup Calculations A.7 Temperature change for water A.8 | Heat calories used to warm water (show A.9 | Heat calories lost by the metal object A.10 Temperature change for metal A.11 Specific heat of the metal (cal/g C) calculation in space below) (show calculation below) Cal/a Specific heat of the meal (J/g °C) Predict the type of metal Show calculations here- Heat-4504|12.0 eafExplanation / Answer
First run
Mass of water = (mass of cup + water) - (mass of cup)
= 52.0 - 14.0
= 38.0 g
Specific Heat capacity of water Cp = 1 cal/g-°C
Temperature change for water = Final temperature - initial temperature
= 36 - 24
= 12 °C
Heat calories used to warm water
= mass of water x Cp x (T2-T1)
= 38 g x 1 cal/g-°C x 12 °C
= 456 cal
Heat calories lost by metal = 456 cal
Temperature change for metal = 95 - 36 = 59°C
Heat lost = mass of metal x Cpm x (T2-T1)
456 cal = 67 g x Cpm x 59°C
Cpm = 0.115 cal/g-°C
Cpm = 0.115 cal/g-°C x 4.184 J/cal = 0.481 J/g-°C
For second run
Mass of water = (mass of cup + water) - (mass of cup)
= 59.0 - 14.0
= 45 g
Specific Heat capacity of water Cp = 1 cal/g-°C
Temperature change for water = Final temperature - initial temperature
= 36 - 24
= 12 °C
Heat calories used to warm water
= mass of water x Cp x (T2-T1)
= 45 g x 1 cal/g-°C x 12 °C
= 540 cal
Heat calories lost by metal = 540 cal
Temperature change for metal = 96 - 36 = 60°C
Heat lost = mass of metal x Cpm x (T2-T1)
540 cal = 67.29 g x Cpm x 60°C
Cpm = 0.134 cal/g-°C
Cpm = 0.134 cal/g-°C x 4.184 J/cal = 0.561 J/g-°C
Average Cpm = (0.561 + 0.481)/2 = 0.521 J/g-°C
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