Question

Steam is passed into $54$g of water at ${30}^o$C till the temperature of the mixture becomes ${90}^o$C. If the latent heat of steam is $536$cal $g^{-1}$, the mass of the mixture will be.

• A. $80$g
• B. $60$g
• C. $50$g
• D. $24$g

Answer 1

The correct option is A $80$g

Let the mass of steam required to raise the temperature of 54 g of water from $30°C$ to $90°C$ be m gram of steam.

Each gram of steam on condensing releases 536 calories of heat. The steam which condenses is at $100°C,$ and it cools to final temperature of $90°C$.

Heat released by m gram of steam on condensing= $536\times m$ calorie

Heat released by m gram of condensed steam condensed to water at $100°C$ to water at $90°C$, the final temperature of the solution= m×specific heat of water× fall of temperature= $m\times 1\times 10=10m$ calories.

Total heat released by steam condensing and then cooling to $90°C=536m+10m=546m$ calories of heat.

Heat required to raise the temperature of 54 g of water at 30°C + m gram of condensed steam from 30°C to 90°C $= (54 + m) ×1×(90°C - 30°C)= (54 + m)×60 calories

Using heat gained = Heat lost

$(54+m)\times 60=546m;==>3240+60m=546m;====>$ $486m=3240,orm=3240/486=80g$ of steam.