5 kg of water at 80 -C is taken in a bucket of negligible heat capacity- 15 kg of water at 20 -C is added to it- What is the temperature of the mixture-

The correct option is C 35 ^∘C Let t^oC be the #160;temperature of mixture .Heat given by hot water at 80^oC #160; =mc(80 t) Heat ta ...

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Byju's Answer

Standard X

Physics

Calorimeter

5 kg of water...

Question

$5 k g$ of water at $80^{\circ} C$ is taken in a bucket of negligible heat capacity, $15 k g$ of water at $20^{\circ} C$ is added to it. What is the temperature of the mixture?

45^{\circ} C

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65^{\circ} C

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85^{\circ} C

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35^{\circ} C

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Solution

The correct option is C $35^{\circ} C$
Let $t^{o} C$ be the temperature of mixture .
Heat given by hot water at $80^{o} C$ $= m c (80 - t)$
Heat taken by cold water at $20^{o} C$ $= m^{'} c (t - 20)$
given , $m = 5 k g, m^{'} = 15 k g$
By the principle of mixtures ,
$5 c (80 - t) = 15 c (t - 20)$
or $80 - t = 3 (t - 20)$
or $4 t = 140$
or $t = 140 / 4 = 35^{o} C$

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5kg of water at 80∘C is taken in a bucket of negligible heat capacity, 15kg of water at 20∘C is added to it. What is the temperature of the mixture?

The correct option is C 35∘CLet toC be the temperature of mixture .Heat given by hot water at 80oC =mc(80−t)Heat taken by cold water at 20oC =m′c(t−20)given , m=5kg,m′=15kgBy the principle of mixtures , 5c(80−t)=15c(t−20)or 80−t=3(t−20)or 4t=140or t=140/4=35oC

$5 k g$ of water at $80^{\circ} C$ is taken in a bucket of negligible heat capacity, $15 k g$ of water at $20^{\circ} C$ is added to it. What is the temperature of the mixture?