Question
Question: What would be the final temperature of a mixture of \(50g\) of water at \({20^0}C\) temperature and ...
What would be the final temperature of a mixture of 50g of water at 200C temperature and 50g of water at 400C?
Solution
Hint
We will use the concept of calorimeter. We can use that Amount of heat radiated by a hot object equals the amount of heat absorbed by the same object at a relatively lower temperature.
⇒Amount of absorption or radiation of heat=m×s×T
Where, m is the mass of the object, s is its specific heat capacity and T is the temperature at which it is placed.
Complete step by step answer
Let the specific heat capacity of water be sw
Given, Mass of water used for both cases, mw=50g
Warmer temperature, T1=400C
Cooler Temperature, T2=200C
Now,
Amount of heat radiated by warm water= m×sw×(T1−T)=50×sw×(40−T)=(40−T)×sw
Amount of heat absorbed by cold water= m×sw×(T−T2)=50×sw×(T−20)=(T−20)×sw
Then,
By the law of calorimeter,
Amount of heat radiated by warm water= Amount of heat absorbed by cold water ⇒(40−T)=(T−20)
After further evaluation, we get the value of the solution, T=300C
We can also simply calculate this value,
As, The same substance in this case water at different temperature is mixed,
Thus, the final mixture temperature will turn out to be the average of the individual temperatures. That means,
T=(400C+200C)/2=300C
Note
The concept of calorimeter is very much essential for the solving of this type of problems. Here, no doubt the value from the calorimeter formula and that after averaging the temperatures are the same, due to the temperatures acting on the same substance. But that is not always the same as there are situations where two different substances are used at different temperatures and we are given to find out the total mixture temperature. Thus, it is more beneficial to use the concept of calorimeter as the chances of uttering an error and getting deviated from the original answer reduces.