If 5 L of water at `50^(@)C` is mixed with 4L of water at `30^(@)C`, what will be the final temperature of water? Take the specific heat capacity of water as `4184 J kg^(-1) K^(-1)`

Explanation

We can use the equation `T_(f)=(m_(1)s_(1)T_(1)+m_(2)s_(2)T_(2))/(m_(1)s_(1)+m_(2)s_(2))`
`m_(1)=5L = 5kg and m_(2)=4L = 4kg, s_(1)=s_(2) and T_(1)=50^(@)C=323K and T_(2)=30^(@)C=303K`
`"So "T_(f)=(m_(1)T_(1)+m_(2)T_(2))/(m_(1)+m_(2))=(5xx323+4xx303)/(5+4)=314.11K`
`T_(f)=314.11K-273K~~41^(@)C`
Suppose if we mix equal number of water `(m_(1)=m_(2))` with `50^(@)C` and `30^(@)c`, then the find temperature is average of two temperatures.
`T_(f)=(T_(1)+T_(2))/(2)=(323+303)/(2)=313K=40^(@)C`
Suppose if both the water are at `30^(@)C` then the final temperature will also `30^(@)C`. It implies that they are at equilibrium and no heat exchange takes place between each other.