A beaker contains 200 gm of water. The heat capacity of the beaker is equal to that of 20 gm of water. The initial temperature of water in the beaker is 20°C. If 440 gm of hot water at 92°C is poured in it, the final temperature (neglecting radiation loss) will be nearest to

To find the final equilibrium temperature of the mixture, we can apply the principle of calorimetry, which states that in an isolated system, the heat lost by the hot substances must equal the heat gained by the cold substances.

First, let us define the given parameters:
Mass of cold water in the beaker, $m_1=200\text{g}$
Water equivalent of the beaker, $W_{\text{beaker}}=20\text{g}$ (since its heat capacity is equal to that of $20\text{g}$ of water)
Initial temperature of the beaker and cold water, $T_1=20\text{°C}$
Mass of hot water added, $m_2=440\text{g}$
Initial temperature of the hot water, $T_2=92\text{°C}$
Let the specific heat capacity of water be $s$.

The total effective mass of the cold system (water + beaker) in terms of water equivalent is:
$m_{\text{cold, total}}=m_1+W_{\text{beaker}}=200\text{g}+20\text{g}=220\text{g}$

Let the final equilibrium temperature of the mixture be $T_f$.

According to the principle of calorimetry:
Heat gained by the cold system = Heat lost by the hot water
$m_{\text{cold, total}}·s·(T_f-T_1)=m_2·s·(T_2-T_f)$

Since the specific heat capacity $s$ is common to both sides, we can cancel it:
$220·(T_f-20)=440·(92-T_f)$

We can simplify the equation by dividing both sides by 220:
$1·(T_f-20)=2·(92-T_f)$

Expand and solve for $T_f$:
$T_f-20=184-2T_f$
$3T_f=184+20$
$3T_f=204$
$T_f=\frac{204}{3}=68\text{°C}$

Therefore, the final equilibrium temperature of the mixture will be 68°C.