Question

Three containers $1,2$ and $3$ have water at different temperatures. The table below shows the final temperature $T$ when different amounts of water (given in liters) are taken from each container and mixed (assume no loss of heat during the process)

$1$	$2$	$3$	$\left({}^{\circ }\mathrm{C}\right)$
$1$	$2$	$-$	$60$
$-$	$1$	$2$	$30$
$2$	$-$	$1$	$60$
$1$	$1$	$1$

The value of (in ℃ to the nearest integer) is

Answer 1

Step 1: Obtain the equation using the table

• According to the table and applying the law of calorimetry, we get,

$\begin{array}{l}1{\mathrm{T}}_1+2{\mathrm{T}}_2=(1+2){60}^{\circ }\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)\\ 1{\mathrm{T}}_2+2{\mathrm{T}}_3=(1+2){30}^{\circ }\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\\ 2{\mathrm{T}}_1+1{\mathrm{T}}_3=(1+2){60}^{\circ }\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)\end{array}$

Step 2: Add all the obtained equation

• The addition of all the obtained equations gives the following,

$\begin{array}{c}3({\mathrm{T}}_1+{\mathrm{T}}_2+{\mathrm{T}}_3)=450\\ ({\mathrm{T}}_1+{\mathrm{T}}_2+{\mathrm{T}}_3)=150\cdot \cdot \cdot \cdot \cdot \cdot \left(4\right)\end{array}$

Step 3: Compute the $\theta$

• It is understood that ${\mathrm{T}}_1+{\mathrm{T}}_2+{\mathrm{T}}_3=(1+1+1)\mathrm{\theta }$. So,

$\begin{array}{l}150=3\mathrm{\theta }\\ \mathrm{\theta }={50}^{\circ }\end{array}$

Hence, the value of nearest integer will be $\mathrm{\theta }={50}^{\circ }$