JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Heat & Thermodynamics

Three containers

C_{1}

C_{2}

and

C_{3}

have water at different temperatures. The table below shows the final temperature T when different amounts of water (given in liters) are taken from each container and mixed (assume no loss of heat during the process)

$C_{1}$	$C_{2}$	$C_{3}$	T
$1 ℓ$	$2 ℓ$	-	$6 0^{\circ} C$
-	$1 ℓ$	$2 ℓ$	$3 0^{\circ} C$
$2 ℓ$	-	$1 ℓ$	$6 0^{\circ} C$
$1 ℓ$	$1 ℓ$	$1 ℓ$	$θ$

The value of

θ

(in

^{\circ} C

to the nearest integer) is ____.

Answer: 50

Solution

Let the three containers hold water at

θ_{1}, θ_{2}, θ_{3}

. With no heat lost, mixing conserves the product of volume and temperature. The first mixing gives

θ_{1} + 2 θ_{2} = 3 (60) = 180

, the second

θ_{2} + 2 θ_{3} = 3 (30) = 90

, and the third

2 θ_{1} + θ_{3} = 3 (60) = 180

. Adding the three,

3 (θ_{1} + θ_{2} + θ_{3}) = 450

, so

θ_{1} + θ_{2} + θ_{3} = 150

. Taking one litre from each gives a mixture at

θ = \frac{θ _{1} + θ _{2} + θ _{3}}{3} = \frac{150}{3} = 5 0^{\circ} C

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Integer answer

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JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Heat & Thermodynamics

Solution

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