2 kg of a liquid at 100 -C is poured in a bucket having the same liquid of mass 8 kg at 12 -C- Find the final temperature of the mixture when its specific heat capacity is 130J kg -1 K -1-

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Standard XII

Chemistry

Units and Dimensions

2 kg of a liq...

Question

$2$ kg of a liquid at $100^{\circ} C$ is poured in a bucket having the same liquid of mass $8$ kg at $12^{\circ} C$ . Find the final temperature of the mixture when its specific heat capacity is $130 J {k g}^{- 1} K^{- 1}$ .

50^{\circ} C

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{24.6}^{\circ} C

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{49.3}^{\circ} C

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{29.6}^{\circ} C

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Solution

The correct option is D ${29.6}^{\circ} C$
Heat gain is equal to heat loss. Let final temperature be $T$ .
So,
$\Rightarrow 2000 \times 1 \times (100 - T) = 8000 \times 1 \times (T - 12) \Rightarrow 200 - 2 T = 8 T - 96 \Rightarrow 10 T = 200 + 96 \Rightarrow T = 29.6 ℃$

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2 kg of a liquid at 100∘C is poured in a bucket having the same liquid of mass 8 kg at 12∘C. Find the final temperature of the mixture when its specific heat capacity is 130Jkg−1K−1.

The correct option is D 29.6∘CHeat gain is equal to heat loss. Let final temperature be T.So,⇒2000×1×(100−T)=8000×1×(T−12)⇒200−2T=8T−96⇒10T=200+96⇒T=29.6℃

$2$ kg of a liquid at $100^{\circ} C$ is poured in a bucket having the same liquid of mass $8$ kg at $12^{\circ} C$ . Find the final temperature of the mixture when its specific heat capacity is $130 J {k g}^{- 1} K^{- 1}$ .

The correct option is D ${29.6}^{\circ} C$
Heat gain is equal to heat loss. Let final temperature be $T$ .
So,
$\Rightarrow 2000 \times 1 \times (100 - T) = 8000 \times 1 \times (T - 12) \Rightarrow 200 - 2 T = 8 T - 96 \Rightarrow 10 T = 200 + 96 \Rightarrow T = 29.6 ℃$