A liquid A of mass 100g at 100° C is added to 50g of a liquid B at temperature 75° C, the temperature of the mixture becomes 90° C . Now if 100g of liquid A at 100° C is added to 50g of liquid B at 50° C, temperature of the mixture will be

Question

A liquid A of mass 100g at 100° C is added to 50g of a liquid B at temperature 75° C, the temperature of the mixture becomes 90° C . Now if 100g of liquid A at 100° C is added to 50g of liquid B at 50° C, temperature of the mixture will be (A) 80° C (B) 60° C (C) 70° C (D) 85° C

Tardigrade Admin · Accepted Answer

Heat loss will always be equal to heat gain.
As per

1^{s t}

condition,

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

As per

2^{nd}

condition,

100 \times S_{A} (100 - θ) = 50 \times S_{B} (θ - 50)

Dividing (ii) by (i), we get

\frac{100 - θ}{100 - 90} = \frac{θ - 50}{90 - 75}

300 - 3 θ = 2 θ - 100

θ = 8 0^{\circ} C

$8 0^{\circ} C$

$6 0^{\circ} C$

$7 0^{\circ} C$

$8 5^{\circ} C$

Q. A liquid A of mass 100g at 100∘C is added to 50g of a liquid B at temperature 75∘C, the temperature of the mixture becomes 90∘C . Now if 100g of liquid A at 100∘C is added to 50g of liquid B at 50∘C, temperature of the mixture will be

80∘C

60∘C

70∘C

85∘C

Heat loss will always be equal to heat gain. As per 1st condition, 100×SA​×(100−90)=50×SB​×(90−75) As per 2nd condition, 100×SA​(100−θ)=50×SB​(θ−50) Dividing (ii) by (i), we get 100−90100−θ​=90−75θ−50​ 300−3θ=2θ−100 θ=80∘C

$8 0^{\circ} C$

$6 0^{\circ} C$

$7 0^{\circ} C$

$8 5^{\circ} C$