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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2019 · 11 January · Evening Shift · Physics · Kinetic Theory of Gases

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

When 100 g100 \mathrm{~g} of a liquid A\mathrm{A} at 100∘C100^{\circ} \mathrm{C} is added to 50 g50 \mathrm{~g} of a liquid BB at temperature 75∘C75^{\circ} \mathrm{C}, the temperature of the mixture becomes 90∘C90^{\circ} \mathrm{C}. The temperature of the mixture, if 100 g100 \mathrm{~g} of liquid AA at 100∘C100^{\circ} \mathrm{C} is added to 50 g50 \mathrm{~g} of liquid B\mathrm{B} at 50∘C50^{\circ} \mathrm{C}, will be:

Options

  1. A85∘C85^{\circ} \mathrm{C}
  2. B60∘C60^{\circ} \mathrm{C}
  3. C80∘C80^{\circ} \mathrm{C}
  4. D70∘C70^{\circ} \mathrm{C}

Correct answer

C. 80∘C80^{\circ} \mathrm{C}

Step-by-step solution

Heat loss = Heat gain =mSΔθ=\mathrm{mS} \Delta \theta So,mASAΔθA=mBSBΔθB\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}} ⇒100×SA×(100−90)=50×SB×(90−75)2 SA=1.5 SB⇒SA=34 SB\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×SA×(100−θ)=50×SB×(θ−50)100 \times \mathrm{S}_{\mathrm{A}} \times(100-\theta)=50 \times \mathrm{S}_{\mathrm{B}} \times(\theta-50) 2×(34)×(100−θ)=(θ−50)2 \times\left(\frac{3}{4}\right) \times(100-\theta)=(\theta-50) 300−3θ=2θ−100400=5θ⇒θ=80∘C\begin{array}{l} 300-3 \theta=2 \theta-100 \\ 400=5 \theta \Rightarrow \theta=80^{\circ} \mathrm{C} \end{array}
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