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

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Heat & Thermodynamics

When 100 g of a liquid A at

10 0^{\circ} C

is added to 50 g of a liquid B at temperature

7 5^{\circ} C

, the temperature of the mixture becomes

9 0^{\circ} C

. The temperature of the mixture, if 100 g of liquid A at

10 0^{\circ} C

is added to 50 g of liquid B at

5 0^{\circ} C

, will be:

Answer: (C)

The first mixing fixes the ratio of the specific heats: heat lost by A equals heat gained by B,

100 S_{A} (100 - 90) = 50 S_{B} (90 - 75)

, i.e.

1000 S_{A} = 750 S_{B}

, so

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

. In the second mixing the same masses meet with B starting at

5 0^{\circ}

100 S_{A} (100 - T) = 50 S_{B} (T - 50)

. Putting

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

75 (100 - T) = 50 (T - 50)

, so

7500 - 75 T = 50 T - 2500

, giving

125 T = 10000

and

T = 8 0^{\circ}

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option