Question

$30g$ of ice at $0^{0}C$ and $20g$ of steam at ${100}^{0}C$ are mixed. The composition of the resultant mixture is:

• A. $40g$ of water and $10g$ steam at ${100}^{0}C$
• B. $10g$ of ice and $40g$ of water at $0^{0}C$
• C. $50g$ of water at ${100}^{0}C$
• D. $35g$ of water and $15g$ of steam at ${100}^{0}C$

Answer 1

Answer: (A)

$40g$ of water and $10g$ steam at ${100}^{0}C$

Energy spent in converting 30 g of ice at $0^{\circ }C$ into 20 g of water at $0^{\circ }C$ $=m_{ice}{L}_{ice}=30\ast 80=2400cal$

Energy released in converting 20g steam at ${100}^{\circ }C$ into 20g of water at ${100}^{\circ }C$ $=m_s{L}_s=20\ast 540=10800cal$

So, all the steam will not be converted into water.

Let $m_s$ amount of stream is converted into this process.

$m_s{L}_S=m_{ice}{L}_{ice}+m_{ice}s\Delta \theta$

$m_s\ast 540=30\ast 80+30\ast 1\ast (100-0)m_s=\frac{2400+3000}{540}{m}_s=\frac{5400}{540}=10g$

So only 10 g steam is converted into ${100}^{\circ }C$ water and whole 30 g of ice is converted to ${100}^{\circ }C$ water.

So at the final point, we have 10 g steam and 40 g water at ${100}^{\circ }C$.

Hence option 'A' is correct.