200 g of ice at -20° C is mixed with 500 g of water at 20° C in an insulating vessel. Final mass of water in vessel is (specific heat of ice =0.5 cal g -10 C -1 )

Question

200 g of ice at -20° C is mixed with 500 g of water at 20° C in an insulating vessel. Final mass of water in vessel is (specific heat of ice =0.5 cal g -10 C -1 ) (A) 700 g (B) 600 g (C) 400 g (D) 200 g

Tardigrade Admin · Accepted Answer

Maximum heat supplied by water Δ Q1 =500 × 1 ×(20-0) =10,000 cal Heat required to raise the temperature of ice upto 0° C Δ Q2 =200 × 0.5 × 20 =2000 cal Δ Q1 -Δ Q2=8000 cal Melts the ice 8000=m × 80 m=100 g So, mass of water is 500+100=600 g.

Q. 200g of ice at −20∘C is mixed with 500g of water at 20∘C in an insulating vessel. Final mass of water in vessel is (specific heat of ice =0.5calg−10C−1 )

700 g

600 g

400 g

200 g

Maximum heat supplied by water ΔQ1​=500×1×(20−0) =10,000cal Heat required to raise the temperature of ice upto 0∘C ΔQ2​=200×0.5×20 =2000cal ΔQ1​−ΔQ2​=8000cal Melts the ice 8000=m×80 m=100g So, mass of water is 500+100=600g.

Q. $200 g$ of ice at $- 2 0^{\circ} C$ is mixed with $500 g$ of water at $2 0^{\circ} C$ in an insulating vessel. Final mass of water in vessel is (specific heat of ice $= 0.5 c a l g^{- 10} C^{- 1}$ )