When \(50 \mathrm{~g}\) of water at \(10^{\circ} \mathrm{C}\) is mixed with \(50 \mathrm{~g}\) of water at…
When \(50 \mathrm{~g}\) of water at \(10^{\circ} \mathrm{C}\) is mixed with \(50 \mathrm{~g}\) of water at \(100^{\circ} \mathrm{C}\). The resultant temperature is
\(80^{\circ} \mathrm{C}\)
\(55^{\circ} \mathrm{C}\)
\(25^{\circ} \mathrm{C}\)
\(45^{\circ} \mathrm{C}\)
Solution
If resulting temperature of the mixture is \(T^{\circ} \mathrm{C}\) then
Heat gained by \(50 \mathrm{~g}\) water at \(10^{\circ} \mathrm{C}\),
\(H_{\text {gain }}=m c \Delta T=50 c(T-10)\)
Heat lost by \(50 \mathrm{~g}\) of water at \(100^{\circ} \mathrm{C}\)
\(H_{\text {loss }}=m c \Delta T=50 c(100-T)\)
According to principle of calorimetry,
Heat gain \(=\) Heat loss
\(\begin{array}{ll}
\Rightarrow & H_{\text {gain }}=H_{\text {loss }} \\
\Rightarrow & 50 c(T-10)=50 c(100-T) \\
\Rightarrow & 2 T=110 \\
& T=55^{\circ} \mathrm{C}
\end{array}\)