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

AI Thread Summary
To convert 1.0 kg of water at 85°C to steam at 118°C, the heat required can be calculated using the formula Q=mcΔT. The specific heat of water is 4186 J/kg°C, and the temperature change involves heating the water to 100°C, converting it to steam, and then heating the steam to 118°C. The latent heat of vaporization for water is approximately 2260 kJ/kg. The discussion highlights confusion regarding the mass of steam produced and the calculation steps involved in determining the total heat required for the phase change and subsequent heating. Clarification on these calculations is sought to resolve the issue.
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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?
 
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If you evaporate 1 kg of water, how much steam do you get?
 
2.26*103kJ/kg?
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