How much heat in joules is needed to raise the temperature

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SUMMARY

The discussion focuses on calculating the heat required to raise the temperature of water, specifically 7.0 L from 0°C to 78.0°C. The correct calculation yields 2.28 x 106 joules using the formula Q = mcΔT, where m is the mass of water (7000 g), c is the specific heat capacity (4.184 J/g°C), and ΔT is the temperature change (78°C). The initial miscalculation of 2.73 x 106 J was corrected after realizing the need to include the factor of 10 in the final answer.

PREREQUISITES
  • Understanding of specific heat capacity
  • Familiarity with the conversion between calories and joules
  • Basic knowledge of temperature change calculations
  • Ability to manipulate scientific notation
NEXT STEPS
  • Learn how to apply the formula Q = mcΔT in different scenarios
  • Explore the conversion methods between calories and joules
  • Investigate the properties of water and its specific heat capacity
  • Practice problems involving heat transfer in various states of matter
USEFUL FOR

Students studying thermodynamics, educators teaching heat transfer concepts, and anyone needing to perform calculations related to thermal energy in water.

comoore
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Homework Statement


How much heat in joules is needed to raise the temperature of 7.0 L of water from 0°C to 78.0°C? (Hint: Recall the original definition of the liter.)

Homework Equations


How much heat in joules is needed to raise the temperature of 7.5 L of water from 0°C to 87.0°C? (Hint: Recall the original definition of the liter.)
2.73e+06 J

The Attempt at a Solution


I answered 2.28^06 J.
I have actually already asked this question on another forum, and they couldn't figure it out either. Please help.
 
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I think so...

I did 7000 gm x 78 C=546000 calories
546000x4.184j/calorie=2284.5 kj
Then converted kj to j
 
OH. MY. GOSH. I didn't put the 10 in when I answered. The answer is correct. 2.28x10^6 J.
 
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comoore said:
7.0 L of water
comoore said:
7.5 L of water
Better turn off the tap.
 
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