Help with understanding how to figure out the amount of heat to boil away water.

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SUMMARY

To calculate the total heat required to boil away 2300 grams of water starting at 17 degrees Celsius, two formulas are essential: Q = c m ΔT for heating the water to boiling point and Qv = m Lv for the phase change from liquid to vapor. The specific heat (c) of water is 1 calorie per gram per degree Celsius, and the heat of vaporization (Lv) is 540 calories per gram. The temperature difference (ΔT) is 83 degrees Celsius, calculated as 100 - 17.

PREREQUISITES
  • Understanding of specific heat capacity
  • Knowledge of heat of vaporization
  • Basic algebra for manipulating equations
  • Familiarity with temperature scales (Celsius)
NEXT STEPS
  • Calculate total heat using Q = c m ΔT and Qv = m Lv
  • Explore the concept of latent heat in phase changes
  • Learn about specific heat capacities of different substances
  • Investigate the implications of heat transfer in thermodynamics
USEFUL FOR

Students studying thermodynamics, physics enthusiasts, and anyone involved in heat transfer calculations.

Mtray524
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1. The problem statement, all variables and given/know

2300 grams of water is heated. If this water starts at a temperature of 17 degrees Celsius, what amount of heat is needed to boil away all of this water? Note that the specific heat of water is 1 calorie per gram per degree Celsius, the Heat of vaporization of water is 540 calories per gram, and water boils at 100 degrees Celsius.

The Attempt at a Solution



I figured that this would be the best formule for me to use Q = c m ΔT, and Qv = m Lv .

Since water boils at 100 degrees celsius, I assumed that I should subtract 100-17 to get 83 degrees as the differnce in temperature.

Can I please get assistance with understanding this probelm.
 
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Actually, you are on the right track. Use your two formulas. The total heat is the sum of the two.
 

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