Thermal Properties of Matter

JEE Main level, single correct, medium

69 of 126

Question 69 of 126

The temperature of equal masses of three different liquids x,yx,y and zz are 10∘C,20∘C10^{\circ}C,20^{\circ}C and 30∘C30^{\circ}C respectively. The temperature of mixture when xx is mixed with yy is 16∘C16^{\circ}C and that when yy is mixed with zz is 26∘C26^{\circ}C. The temperature of mixture when xx and zz are mixed will be :

Show answer and solution

Answer: (D) 23.84∘C23.84^{\circ}C

When A&B are mixed
m1S1(16−10)=m2S2(20−16){m}_{1}{S}_{1}(16-10)={m}_{2}{S}_{2}(20-16)
6m1S1=4m2S2.......(1)6{m}_{1}{S}_{1}=4{m}_{2}{S}_{2}.......(1)
When
B&C are mixed
m2S2(26−20)=m3S3(30−26){m}_{2}{S}_{2}(26-20)={m}_{3}{S}_{3}(30-26)
6m2S2=4m3S3...........(2)6{m}_{2}{S}_{2}=4{m}_{3}{S}_{3}...........(2)
From equation
(1)(1) and (2)(2)
9m1S1=4m3S39{m}_{1}{S}_{1}=4{m}_{3}{S}_{3}
When
A&C are mixed
m1S1(T−10)=m3S3(30−T){m}_{1}{S}_{1}(T-10)={m}_{3}{S}_{3}(30-T)
T=30m3S3+10m1S1m1S1+m3S3T=\dfrac{30{m}_{3}{S}_{3}+10{m}_{1}{S}_{1}}{{m}_{1}{S}_{1}+{m}_{3}{S}_{3}}
T=10(3m3S3+m1S1)m3S3+m1S1T=\dfrac{10\left(3{m}_{3}{S}_{3}+{m}_{1}{S}_{1}\right)}{{m}_{3}{S}_{3}+{m}_{1}{S}_{1}}
T=10(3m3S3+49m3S3)m3S3+49m3S3T=\dfrac{10\left(3{m}_{3}{S}_{3}+\dfrac{4}{9}{m}_{3}{S}_{3}\right)}{{m}_{3}{S}_{3}+\dfrac{4}{9}{m}_{3}{S}_{3}}
T=10×3113T=\dfrac{10\times 31}{13}
=23.84∘C=23.84^{\circ}C

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Thermal Properties of Matter (Calorimetry), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.