Question

Three containers \(C_1\), \(C_2\) and \(C_3\) have water at different temperatures. The table below shows the final temperature \(T\) when different amounts of water (given in litres) are taken from each containers and mixed (assume no loss of heat during the process)

\(C_1\)	\(C_2\)	\(C_3\)	\(T\)
1 l	2 l	-	\(60^\circ\)C
-	1 l	2 l	\(30^\circ\)C
2 l	-	1 l	\(60^\circ\)C
1 l	1 l	1 l	\(\theta\)

The value of \(\theta\) (in \(^\circ\)C to the nearest integer) is ..........

Hint:

Same liquid, so the mixing temperature is the volume-weighted mean of the initial temperatures. Form three equations in \(T_1,T_2,T_3\).

Answer 1

Correct Answer: 50

Let initial temperatures be \(T_1,T_2,T_3\). Mixing water: final temperature \(=\) weighted mean (same specific heat and density).
Row 1: \(\frac{T_1+2T_2}{3}=60\Rightarrow T_1+2T_2=180\).
Row 2: \(\frac{T_2+2T_3}{3}=30\Rightarrow T_2+2T_3=90\).
Row 3: \(\frac{2T_1+T_3}{3}=60\Rightarrow2T_1+T_3=180\).
From row 1: \(T_1=180-2T_2\). Row 3: \(360-4T_2+T_3=180\Rightarrow T_3=4T_2-180\). Row 2: \(T_2+8T_2-360=90\Rightarrow T_2=50^\circ\)C, \(T_1=80^\circ\)C, \(T_3=20^\circ\)C.
\(\theta=\frac{80+50+20}{3}=50^\circ\)C.
Answer: 50