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A thermos flask contains $250 \mathrm{~g}$ of coffee at $90^{\circ} \mathrm{C}$. To this $20 \mathrm{~g}$ of milk at $5^{\circ} \mathrm{C}$ is added. After equilibrium is established, the temperature of the liquid is
(Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as $\left.1.00 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\right)$
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(Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as $\left.1.00 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\right)$
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2685 Upvotes
Verified Answer
The correct answer is:
$83.7^{\circ} \mathrm{C}$
Let after getting equilibrium stage final temperature is $T$.
Then, according to loss of calorimetry
Heat loss $=$ Heat gain
$$
\begin{aligned}
\Rightarrow(250)(1)(90-T) & =20(1)(T-5) \\
\Rightarrow \quad T & =83.7^{\circ} \mathrm{C}
\end{aligned}
$$
Then, according to loss of calorimetry
Heat loss $=$ Heat gain
$$
\begin{aligned}
\Rightarrow(250)(1)(90-T) & =20(1)(T-5) \\
\Rightarrow \quad T & =83.7^{\circ} \mathrm{C}
\end{aligned}
$$
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