Question

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

Answer 1

This question can be solved by using the concept of calorimetry. According to the principle of calorimetry, the heat lost by a hot body is equal to the heat that the cold body gains. By using this principle we can equate the two bodies ’ heat and thus find out the final temperature. We can also see that in this case both the bodies are water and thus the heat capacities are the same.

Complete answer:
We know that the heat of a body can be represented as:
$\Rightarrow Q = m \times C \times \Delta T$
Here m is the mass of the system.
C is the heat capacity of the system which is 1 calorie for water.
$\Delta T = {T_{final}} - {T_{initial}}$ is the change in temperature.
The water equivalent of the water + beaker system initially is 220g.
Thus we can say that the heat gained by the cold body is:
$\Rightarrow Q = 220 \times 1 \times (T - 20)$
The heat lost by the hot body is:
$\Rightarrow Q = 440 \times 1 \times (92 - T)$
Now we can equate the heat of the hot body to that of the cold body.
$\Rightarrow 220 \times 1 \times (T - 20) = 440 \times 1 \times (92 - T)$
$\Rightarrow 220T - 4400 = 40480 - 440T$
$\Rightarrow 660T = 44880$
$\Rightarrow T = {68^o}C$

Thus we can say that the correct answer for this question is option (B).

Note:
While substituting the temperature we have to do it in such a way that the value of $\Delta T$ is always positive. So for a cold body, the final temperature is unknown and for a hot body, the initial temperature is unknown. This is because the hot body will lower its temperature and the cold body will increase its temperature after heat exchange.