Question

1 kg ice at - 10${}^{0}C$ is mixed with 1kg water at ${50}^{0}C$. Then correct options are :

• A. equilibrium temperature = $0^{0}C$
• B. mass of water in mixture $[1+\frac{45}{80}]$kg
• C. mass of ice in mixture $[1-\frac{45}{80}]$kg
• D. mass of ice in mixture = 1kg

Answer 1

Answer: (A), (B), (C)

The correct options are
A equilibrium temperature = $0^{0}C$
B mass of water in mixture $[1+\frac{45}{80}]$kg
C mass of ice in mixture $[1-\frac{45}{80}]$kg

Heat taken by ice = 5 Kcal + 80 Kcal = 85 Kcal
Heat given by water = 1 1 50 = 50 Kcal
Heat taken > Heat given so, ice will not complete melt let m g ice melt then
1 x 1/2 x 10 + 80 m = 50
80 m = 45 $⟹m=\frac{45}{80}$