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Question: Answered & Verified by Expert
$0.1 \mathrm{~m}^{3}$ of water at $80^{\circ} \mathrm{C}$ is mixed with $0.3 \mathrm{~m}^{3}$ of water at $60^{\circ} \mathrm{C}$. The final temperature of the mixture is
PhysicsThermal Properties of MatterKCETKCET 2009
Options:
  • A $65^{\circ} \mathrm{C}$
  • B $70^{\circ} \mathrm{C}$
  • C $60^{\circ} \mathrm{C}$
  • D $75^{\circ} \mathrm{C}$
Solution:
2626 Upvotes Verified Answer
The correct answer is: $65^{\circ} \mathrm{C}$
Let the final temperature of the mixture be $t$
Heat lost by water at $80^{\circ} \mathrm{C}$
$\begin{aligned}
&=\mathrm{ms} \Delta \mathrm{t} \\
&=0.1 \times 10^{3} \times \mathrm{s}_{\text {water }} \times\left(80^{\circ}-\mathrm{t}\right) \\
&\quad\left(\because \mathrm{m}=\mathrm{V} \times \mathrm{d}=0.1 \times 10^{3} \mathrm{~kg}\right)
\end{aligned}$
Heat gained by water at $60^{\circ} \mathrm{C}$
$=0.3 \times 10^{3} \times \mathrm{s}_{\text {water }} \times\left(\mathrm{t}-60^{\circ}\right)$
According to principle of calorimetry
Heat lost $=$ Heat gained
$\therefore \quad 0.1 \times 10^{3} \times \mathrm{s}_{\text {water }} \times\left(80^{\circ}-\mathrm{t}\right)$
$=0.3 \times 10^{3} \times s_{\text {water }} \times\left(t-60^{\circ}\right)$
or $\quad\left(80^{\circ}-\mathrm{t}\right)=3 \times\left(\mathrm{t}-60^{\circ}\right)$
or $\quad 4 \mathrm{t}=260^{\circ}$
or $\quad \mathrm{t}=65^{\circ} \mathrm{C}$

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