Question

5g of ice at $0^{o}C$ is mixed with 5 g of steam at ${100}^{o}C,$ what is the final temperature?

• A. ${100}^{o}C$
• B. ${50}^{o}C$
• C. $0^{o}C$
• D. None of these

Answer 1

Answer: (D)

. ${100}^{o}C$.

The heat energy required for 5gm of ice at $0^o$ C to get converted to water at ${100}^o$ C is latent heat required + heat required to change its temperature by ${100}^o$ C = $mL+ms△T=(80\cdot 5)+(5\cdot 1\cdot 100)=900calories$.

Now, the heat liberated by 5gm of steam at ${100}^o$ C to get converted to water at ${100}^o$ C is 5⋅537 = 2685 calories.

So, heat energy is enough for 5gm of ice to get converted to 5gm of water at ${100}^o$ C.

So, only 900 calories of heat energy will be liberated by steam, so the amount of steam that will be converted to water at the same temperature is 900/537 = 1.66 g.

So, the final temperature of the mixture will be ${100}^o$ C.