Steam at 100 degrees Celsius is bubbled into 250 cm(adsbygoogle = window.adsbygoogle || []).push({}); ^{3}of water at room temperature in a calorimeter cup. How much steam will have been added when the water in the cup is at 60 degrees Celsius? (Ignore the effect of the cup.)

So I did:

-(Heat lost by steam) = Heat gained by water

m_{steam}L_{vap}+ m_{steam}c(40[degrees]C) = m_{water}c(40[degrees]C), where c = specific heat of water

solving for m_{steam}, i get 1.72* 10^-2 kg.

But the answer is 1.79* 10^-2 kg. What's wrong with my calculation?

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# Homework Help: Specific and latent heat question

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