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Standard XII

Chemistry

Relating Translational KE and T

When 100 g of...

Question

When 100 g of a liquid A at $100^{0}$ C is added to 50 g of a liquid B at temperature $75^{0}$ C, the temperature of the mixture becomes $90^{0}$ C. The temperature of the mixture, if 100 g of liquid A at $100^{0}$ C is added to 50 g of liquid B at $50^{0}$ C, will be:

60^{0}

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70^{0}

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85^{0}

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80^{0}

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Solution

The correct option is D $80^{0}$ C

To find final temperature of mixture:

$100 \times S_{A} \times [100 - 90] = 50 \times S_{B} \times (90 - 75)$

$2 S_{A} = 1.5 S_{B}$

$S_{A} = \frac{3}{4} S_{B}$

Now, $100 \times S_{A} \times [100 - T] = 50 \times S_{B} (T = 50)$

$2 \times (\frac{3}{4}) (100 - T) = (T - 50)$

$300 - 3 T = 2 T - 100$

$400 = 5 T$

$T = 80$

Hence option D is correct

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When 100 g of a liquid A at 1000C is added to 50 g of a liquid B at temperature 750C, the temperature of the mixture becomes 900C. The temperature of the mixture, if 100 g of liquid A at 1000C is added to 50 g of liquid B at 500C, will be:

The correct option is D 800CTo find final temperature of mixture:100×SA×[100−90]=50×SB×(90−75)2 SA=1.5 SBSA=34 SBNow, 100×SA×[100−T]=50×SB(T=50)2×(34)(100−T)=(T−50)300−3 T=2 T−100400=5 T T=80Hence option D is correct

When 100 g of a liquid A at $100^{0}$ C is added to 50 g of a liquid B at temperature $75^{0}$ C, the temperature of the mixture becomes $90^{0}$ C. The temperature of the mixture, if 100 g of liquid A at $100^{0}$ C is added to 50 g of liquid B at $50^{0}$ C, will be: