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                                    Steam is passes into 22 g of water at `20^@C`. The mass of water that will be present when the water acquires a temperature of `90^@C` (Latent heat of steam is `540 cal//g`) isA. 24.8 gB. 24 gC. 36.6 gD. 30 g | 
                            
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Answer» Correct Answer - A `mL_S+ms_(omega)(100-90)=22S_omega(90-20)` `mxx540+mxx1xx10=22xx70` `m=(22xx70)/(550)=2.8g` Total mass of water `=22+2.8=24.8g`  | 
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