Question

$200g$ of ice at $-{20}^{o}C$ is mixed with $500g$ of water at ${20}^{o}C$ in an insulating vessel. Final mass of water in vessel is (specific heat of ice =$0.5cal{g}^{-1o}{C}^{-1})$

• A. $700g$
• B. $600g$
• C. $400g$
• D. $200g$

Answer 1

The correct option is B $600g$

Given :

200g ice at -20${}^{o}C$

500g water at 20${}^{o}C$

$Q=ms\Delta T$

Solution :

We will bring both condition at that 0${}^{o}C$ water

Case 1

Heat generated in bringing 200g of ice at -20${}^{o}C$ to0${}^{o}C$ of water

$-Q_1=200\times 0.5\times 20+200\times 80$

$Q_1=-18000$cal

Case 2

Heat generated in bringing 500g of water at 20${}^{o}C$ to 0${}^{o}C$ of water

$Q_2=500\times 1\times 20$

$Q_2=10000$cal

Net heat we hav now is $Q_{net}=Q_1+Q_2$

$Q_{net}=-8000cal$

Total water at $0^{o}C$ is 700g

left heat is in negative so it will be used to convert water in ice.

so $mL=8000$

$m\times 80=8000$

m=100g it is mass of ice left in vessel

mass of water left is 700g-100g=600g

The Correct Opt=B