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Steam at 100° C is passed into 20 g of water at 10° C. When water acquires a temperature of 80° C, the mass of water present will be: [Take specific heat of water =1 cal g−1C−1 and latent heat of steam =540 cal g−1] |
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Answer» Steam at 100° C is passed into 20 g of water at 10° C. When water acquires a temperature of 80° C, the mass of water present will be: [Take specific heat of water =1 cal g−1C−1 and latent heat of steam =540 cal g−1] |
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