11. A temperature of a 50 kg block increases by 15°C when 337,500 J of thermal energy are added to the block. (Chapter 5 – Pages 141-142) a. What is the specific heat of the object? Show the appropriate equation from your book and show your work with units. b. What is the block made of? Use the chart on page 141. c. Is this block a good material for insulators or conductors?
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Heat = mass × specific heat × temperature change
q = m C ΔT
Given:
q = 337500 J
m = 50 kg
ΔT = 15°C
Substitute:
337500 J = (50 kg) C (15°C)
C = 450 J/kg/°C
Specific heat is usually recorded in J/g/°C or kJ/kg/°C. Converting:
C = 0.45 J/g/°C = 0.45 kJ/kg/°C
q = m C ΔT
Given:
q = 337500 J
m = 50 kg
ΔT = 15°C
Substitute:
337500 J = (50 kg) C (15°C)
C = 450 J/kg/°C
Specific heat is usually recorded in J/g/°C or kJ/kg/°C. Converting:
C = 0.45 J/g/°C = 0.45 kJ/kg/°C
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