Steam is passed into 56 g of water at 30∘C till the temperature of mixture becomes 90∘C. If the latent heat of condensation of steam is 536 cal/g, the mass of the mixture will be:(Specific heat capacity of water is 1 cal/g∘C)

Steam is passed into 56 g of water at 30∘C till the temperature of m

1.	Steam is passed into 56 g of water at 30∘C till the temperature of mixture becomes 90∘C. If the latent heat of condensation of steam is 536 cal/g, the mass of the mixture will be:(Specific heat capacity of water is 1 cal/g∘C)
Answer» Steam is passed into $56 g$ of water at $30^{\circ} C$ till the temperature of mixture becomes $90^{\circ} C$ . If the latent heat of condensation of steam is $536 cal/g$ , the mass of the mixture will be: (Specific heat capacity of water is $1 {cal/g}^{\circ} C$ )

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Steam is passed into 56 g of water at 30∘C till the temperature of mixture becomes 90∘C. If the latent heat of condensation of steam is 536 cal/g, the mass of the mixture will be:(Specific heat capacity of water is 1 cal/g∘C)

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