Question

heat energy required to convert 1 gram steam at 100°C to water at 50°C. Explain longly. but fastly I will mark him brainliest​

Answer 1

Answer:

We know specific heat of ice is 0.5 cal/gK, specific heat of water=1cal/gK

Also latent heat of fusion= 80 cal/g

latent heat of vaporization=540 cal/g

The following processes occur during the above said conversion.

Ice(263K)→ice(273K)→water(273K)→water(373K)→steam(373K)

Thus total heat absorbed=(0.5×10+80+1×100+540) cal=725 cal=3045J

Hope this helps you....

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Answer 2

Answer:

716 calories

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