Question

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

• A. 70 ${}^{∘}$ C
• B. 60${}^{∘}$C
• C. 75 ${}^{∘}$C
• D. 65${}^{∘}$C

Answer 1

Answer: (D)

Let the final temperature of the mixture be $t$
Heat lost by water at $80^{∘}C$
$=msΔt$
$=0.1×10^3×s_{\text{water }}×\left(80^{∘}−t\right)$
$\left(âˆ\mu m=V×d=0.1×10^{3}kg\right)$
Heat gained by water at $60^{∘}C$
$=0.3×10^3×s_{\text{water }}×\left(t−60^{∘}\right)$
According to principle of calorimetry Heat lost = Heat gained
$∴0.1×10^3×s_{\text{water }}×\left(80^{∘}−t\right)$
$=0.3×10^3×s_{\text{water }}×\left(t−60^{∘}\right)$
or $\left(80^{∘}−t\right)=3×\left(t−60^{∘}\right)$
or $4t=260^{∘}$
or $t=65^{∘}C$