2 kg water at 100°C and 2.5 kg water at 50°C is kept in two identical containers A and B respectively of water equivalent 0.5 kg. If water of container A is poured into container B the final temperatu

Question

2 kg water at 100°C and 2.5 kg water at 50°C is kept in two identical containers A and B respectively of water equivalent 0.5 kg. If water of container A is poured into container B the final temperature of mixture is $T_{1}$ and if water of container B is poured into container A the final temperature is $T_{2}$ . (heat loss is negligible)

TestHub · Accepted Answer

A's water poured into B (container B at 50°C): $T_{1} = \frac{2 ( 100 ) + ( 2.5 + 0.5 ) ( 50 )}{2 + 3} = \frac{350}{5} = 7 0^{\circ} C$ .

B's water poured into A (container A at 100°C): $T_{2} = \frac{2.5 ( 50 ) + ( 2 + 0.5 ) ( 100 )}{2.5 + 2.5} = \frac{375}{5} = 7 5^{\circ} C$ .

Then $15 T_{1} = 1050 = 14 T_{2}$ and $T_{1} = 7 0^{\circ} C$ . Answer: (A), (C)

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