JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 24)
When 100 g of a liquid A at 100oC is added to 50 g of a liquid B at temperature 75oC, the temperature of the mixture becomes 90oC. The temperature of the mixture, if 100 g of liquid A at 100oC is added to 50 g of liquid B at 50oC, will be :
60oC
70oC
85oC
80oC
Explanation
100 $$ \times $$ SA $$ \times $$ [100 $$-$$ 90] = 50 $$ \times $$ SB $$ \times $$ (90 $$-$$ 75)
2SA = 1.5 SB
SA = $${3 \over 4}$$SB
Now, 100 $$ \times $$ SA $$ \times $$ [100 $$-$$ T] = 50 $$ \times $$ SB (T $$-$$ 50)
2 $$ \times $$ $$\left( {{3 \over 4}} \right)$$ (100 $$-$$ T) = (T $$-$$ 50)
300 $$-$$ 3T = 2T $$-$$ 100
400 = 5T
T = 80
2SA = 1.5 SB
SA = $${3 \over 4}$$SB
Now, 100 $$ \times $$ SA $$ \times $$ [100 $$-$$ T] = 50 $$ \times $$ SB (T $$-$$ 50)
2 $$ \times $$ $$\left( {{3 \over 4}} \right)$$ (100 $$-$$ T) = (T $$-$$ 50)
300 $$-$$ 3T = 2T $$-$$ 100
400 = 5T
T = 80
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