When 100 ~g of a liquid A at 100^ ^ ^ C is added to 50 ~g of a liquid B at temperature 75^ ^ ^ C , the temperature of the mixture becomes 90^ ^ ^ C . The temperature of the mixture, if 100 ~g of liquid A at 100^ ^ ^ C

Correct answer: C. 80^ ^ ^ C Heat loss = Heat gain =mS Δ , m_ A S_ A Δ θ_ A =m_ B S_ B Δ θ_ B array l 100 × S_ A ×(100-90)=50 × S_ B ×(90-75) 2 ~S_ A =1.5 ~S_ B S_ A =(3)/(4) ~S_ B array Now, 100 × S_ A ×( 100- θ)= 50 × S_ B ×(θ- 50)2 ×((3)/(4)) ×( 100- θ)=(θ-50) array l 300-3 θ=2 θ-100 400=5 θ θ=80^ ^ ^ C array

JEE Main

Exam: JEE Main
Subject: Physics
Chapter: Kinetic Theory of Gases

JEE Main · Physics · Kinetic Theory of Gases

When 100 ~g of a liquid A at 100^ ^ ^ C is added to 50 ~g of a...

When

100 \mathrm{~g}

of a liquid

\mathrm{A}

100^{{}^{{}^{\circ}}} \mathrm{C}

is added to

50 \mathrm{~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 \mathrm{~g}

of liquid

A

100^{{}^{{}^{\circ}}} \mathrm{C}

is added to

50 \mathrm{~g}

of liquid

\mathrm{B}

50^{{}^{{}^{\circ}}} \mathrm{C}

, will be:

Options

A $85^{{}^{{}^{\circ}}} \mathrm{C}$
B $60^{{}^{{}^{\circ}}} \mathrm{C}$
C $80^{{}^{{}^{\circ}}} \mathrm{C}$
D $70^{{}^{{}^{\circ}}} \mathrm{C}$

Correct answer

C. $80^{{}^{{}^{\circ}}} \mathrm{C}$

Step-by-step solution

Heat loss = Heat gain

=\mathrm{mS} \Delta \theta\mathrm{So}, \mathrm{m}_{\mathrm{A}} \mathrm{S}_{\mathrm{A}} \Delta \theta_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \mathrm{S}_{\mathrm{B}} \Delta \theta_{\mathrm{B}}

\begin{array}{l} \Rightarrow 100 \times \mathrm{S}_{\mathrm{A}} \times(100-90)=50 \times \mathrm{S}_{\mathrm{B}} \times(90-75) \\ 2 \mathrm{~S}_{\mathrm{A}}=1.5 \mathrm{~S}_{\mathrm{B}} \Rightarrow \mathrm{S}_{\mathrm{A}}=\frac{3}{4} \mathrm{~S}_{\mathrm{B}} \end{array}

Now,

100

\times \mathrm{S}_{\mathrm{A}} \times(

100-

\theta)=

\times \mathrm{S}_{\mathrm{B}} \times(\theta-

50)2

\times\left(\frac{3}{4}\right) \times(

100-

\theta)=(\theta-50)

\begin{array}{l}300-3 \theta=2 \theta-100 \\ 400=5 \theta \Rightarrow \theta=80^{{}^{{}^{\circ}}} \mathrm{C} \end{array}

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