Question

$0.1\,m^{3}$ of water at $80\,^\circ$C is mixed with $0.3\,m^3$ of water at $60\,^\circ$C. The final temperature of the mixture is

• A. 70 $^\circ$ C
• B. 60$^\circ$C
• C. 75 $^\circ$C
• D. 65$^\circ$C

Answer 1

Answer: (D)

65$^\circ$C

Answer 2

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