How much heat is required to convert 1.0 kg of water at 85°C to steam at 118°C?

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SUMMARY

The discussion centers on calculating the heat required to convert 1.0 kg of water at 85°C to steam at 118°C. The relevant formula used is Q = mcΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature. The specific heat capacity of water is 4186 J/kg°C. The total heat required includes both the heat needed to raise the temperature of the water to 100°C and the heat required for the phase change from water to steam, followed by heating the steam to 118°C.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with the specific heat capacity of water (4186 J/kg°C).
  • Knowledge of phase changes and latent heat of vaporization (2.26 x 10³ kJ/kg).
  • Ability to apply the formula Q = mcΔT in practical scenarios.
NEXT STEPS
  • Calculate the heat required to raise the temperature of water from 85°C to 100°C using Q = mcΔT.
  • Determine the latent heat required for the phase change from water to steam at 100°C.
  • Calculate the heat needed to raise the temperature of steam from 100°C to 118°C.
  • Combine all calculated heat values to find the total heat required for the conversion.
USEFUL FOR

Students studying thermodynamics, physics educators, and anyone involved in heat transfer calculations or related engineering fields.

kee23
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Homework Statement



How much heat is required to convert 1.0 kg of water at 85°C to steam at 118°C?

Homework Equations



Q=mc(delta)T
Qw+Qs=0?

The Attempt at a Solution


we know the m(water)=1.0kg and c(water)=4186J/kgC and T=85C
so using Heat (Q) formula Q=mcT--- (1.0kg)(4186J/kgC)(85C)=306*105
then from here where I got stuck because there is no mass for steam and I thought of finding the mass steam first, so can someone show me the step?
 
Physics news on Phys.org
If you evaporate 1 kg of water, how much steam do you get?
 
2.26*103kJ/kg?
I don't get your question...
 

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