Question

A beaker contains $200gm$ of water. The heat capacity of the beaker is equal to that of $20gm$ of water. The initial temperature of water in the beaker is ${20}^{\circ }C$ . If $440gm$ of hot water at ${92}^{\circ }C$ is poured in it, the final temperature, neglecting radiation loss, will be nearest to:
(A) ${58}^{\circ }C$
(B) ${68}^{\circ }C$
(C) ${73}^{\circ }C$
(D) ${78}^{\circ }C$

Answer 1

Hint: In order to answer this question, first we will assume the final temperature be $T$ . And then we will find the heat absorbed by the cold water and beaker system and also we will conclude the heat lost by the hot water. And finally we will apply the conservation of heat energy to find our final temperature.

Complete answer:
As per the question, given that-
Mass of cold water, $M_c=200g$
Mass of hot water, $M_h=440g$
As we know that, the specific heat of water is, $S_w=1cal/g^{\circ }C$
Now, initial temperature of water in the beaker:
$T_c=T_b={20}^{\circ }C$
where, $T_c$ is the temperature of cold water, and
$T_b$ is the temperature of the beaker.
And, $T_h={92}^{\circ }C$
where, $T_h$ is the temperature of hot water.
Now, we have to find the final temperature, i.e.. $T=?$
As we know the formula of Heat capacity:
$\therefore C=mS$
where, $C$ is the heat capacity,
$m$ is the mass and $S$ is the specific heat.
So, for $20gm$ of water, the heat capacity will be:
$$\begin{gathered} \therefore C=20\times 1cal/g^{\circ }C \\ =20cal{/}^{\circ }C \end{gathered}$$
Or, the heat capacity of the beaker:
$C_{bea\ker }=20cal{/}^{\circ }C=m_b\times S_b$
Now, according to the principle of Calorimetry:-
$$\begin{gathered} \therefore Heatlostbyhotbody=Heatgainedbycoldbody \\ \Rightarrow Q_h=Q_c+Q_{bea\ker } \end{gathered}$$
$$\begin{gathered} \Rightarrow m_h\times S_h\times (\mathrm{\Delta }T{)}_h=m_c\times S_c\times (\mathrm{\Delta }T{)}_c+m_{bea\ker }\times S_{bea\ker }\times (\mathrm{\Delta }T{)}_{bea\ker } \\ \Rightarrow 440\times 1\times (92-T)=200\times 1\times (T-20)+m_{bea\ker }\times S_{bea\ker }\times (T-20) \end{gathered}$$
As we can see above, $m_{bea\ker }\times S_{bea\ker }=C_{bea\ker }$ , we will put the value of $C_{bea\ker }$ instead of $m_{bea\ker }\times S_{bea\ker }$ :
$$\begin{gathered} \Rightarrow 440(92-T)=(T-20)(200+20) \\ \Rightarrow 184-2T=T-20 \\ \Rightarrow 3T=204 \\ \Rightarrow T={68}^{\circ }C \end{gathered}$$
Therefore, the final temperature, neglecting radiation loss, will be nearest to ${68}^{\circ }C$ .
Hence, the correct option is (B) ${68}^{\circ }C$ .

Note:
The law of energy conservation is a crucial law in thermodynamics. It asserts that energy cannot be created or destroyed. However, we can change it from one form to another. When all forms of energy are taken into account, the overall energy of an isolated system remains constant.