Termodynamics: heat supplied with loss of mass

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SUMMARY

The discussion focuses on calculating the mass of liquid lead required to heat 1 kg of liquid water from 20ºC to 60ºC, while accounting for a loss of 100 mg of water due to vaporization at 1 atm pressure. The participant confirms that the final mass of water is 1 kg minus the 100 mg loss, and emphasizes the importance of including the heat of vaporization in the heat balance calculation. The correct approach involves using the equation Q = m*(delta)H to determine the heat transfer associated with the phase change.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with phase change equations and heat of vaporization.
  • Basic knowledge of mass and energy conservation in thermodynamic systems.
  • Ability to perform calculations involving specific heat capacities and temperature changes.
NEXT STEPS
  • Study the heat transfer equations in thermodynamics, particularly Q = m*(delta)H.
  • Learn about the heat of vaporization for various substances, including water.
  • Explore the concept of mass loss during phase changes and its impact on thermal calculations.
  • Investigate the effects of pressure on boiling points and vaporization in thermodynamic systems.
USEFUL FOR

Students studying thermodynamics, engineers working with heat transfer systems, and anyone involved in thermal management or energy calculations in physical sciences.

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Homework Statement



Find the mass of liquid lead at fussion temperature needed to heat up 1kg of liquid water from 20ºC to 60ºC, considering P = 1 atm, and a loss of 100 mg of water due to local vaporization.

My question is: what do I do with that loss of mass? Do I calculate Q for m = (1000 - 0,1) g?

I'm really at loss with that.

Thanks.
 
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Final mass of water is 1kg-100mg, no doubt about it. Question is whether heat of vaporization should be taken into account or not in heat balance. However, check if it matters - we are talking about 10-4 of the initial mass, most likely difference will be way below accuracy of the final result.

--
 
Yes, apparently I had to use that mass to calculate heat through the change of phase equation: Q = m*(delta)H.

Thanks.
 

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