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$0.1 \mathrm{~m}^3$ 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
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The correct answer is:
$65^{\circ} \mathrm{C}$
Let the final temperature of mixture be $t$.
Heat lost by water at $80^{\circ} \mathrm{C}$
$=m s \Delta t$
$=0.1 \times 10^3 \times s_{\text {water }} \times\left(80^{\circ}-t\right)$
$\left(\because m=V \times d=0.1 \times 10^3 \mathrm{~kg}\right)$
Heat gained by water at $60^{\circ} \mathrm{C}$
$=0.3 \times 10^3 \times s_{\text {water }} \times\left(t-60^{\circ}\right)$
According to principle of calorimetry
Heat lost $=$ Heat gained
$\therefore \quad 0.1 \times 10^3 \times s_{\text {water }} \times\left(80^{\circ}-t\right)$
$=0.3 \times 10^3 \times s_{\text {water }} \times\left(t-60^{\circ}\right)$
or $\quad\left(80^{\circ}-t\right)=3 \times\left(t-60^{\circ}\right)$
or $\quad 4 t=260^{\circ}$
or $\quad t=65^{\circ} \mathrm{C}$
Heat lost by water at $80^{\circ} \mathrm{C}$
$=m s \Delta t$
$=0.1 \times 10^3 \times s_{\text {water }} \times\left(80^{\circ}-t\right)$
$\left(\because m=V \times d=0.1 \times 10^3 \mathrm{~kg}\right)$
Heat gained by water at $60^{\circ} \mathrm{C}$
$=0.3 \times 10^3 \times s_{\text {water }} \times\left(t-60^{\circ}\right)$
According to principle of calorimetry
Heat lost $=$ Heat gained
$\therefore \quad 0.1 \times 10^3 \times s_{\text {water }} \times\left(80^{\circ}-t\right)$
$=0.3 \times 10^3 \times s_{\text {water }} \times\left(t-60^{\circ}\right)$
or $\quad\left(80^{\circ}-t\right)=3 \times\left(t-60^{\circ}\right)$
or $\quad 4 t=260^{\circ}$
or $\quad t=65^{\circ} \mathrm{C}$
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