Question

40 g of hot water is poured in 100 g of cold water, when the temperature of cold water rises by 10$^\circ $C. If the temperature of hot water is 60$^\circ $C, calculate the initial temperature of cold water.

Answer 1

Hint:We have the masses and upon mixing their final temperatures will be the same. By writing the expression for heat energy lost and gained by hot and cold water respectively, and then equating them at the equilibrium, we can get the required answer.

Formula used:
The heat energy absorbed or released by a substance is given in terms of its mass, its specific heat capacity and change in temperature by the following expression.
$Q = mc\Delta T$

Complete answer:
Let us first consider the case of hot water. The mass of the hot water is given as
\[M = 40g\]
Let c be the specific heat capacity of water. Its value for water is
$c = 4.2J{\text{ }}{g^{ - 1}}{\text{ }}^\circ {C^{ - 1}}$
The initial temperature of the hot water is given as
Initial temperature $ = 60^\circ C$
Let the final temperature of water be $x^\circ C$.
Now we will consider the case of cold water. The mass of the cold water is given as
$m = 100g$
The specific heat capacity of cold water is the same as hot water as the material is the same. We don’t know the initial temperature of cold water but its final temperature is the same as the final temperature of the hot water which is $x^\circ C$. But we are given the change in temperature for cold water which is
$\Delta {T_C} = 10^\circ C$
When mixed with each other, the hot water will give out the following amount of heat.
$ = Mc\Delta {T_H} = 40 \times 4.2 \times (60 - x)$
Similarly, the heat absorbed by the cold water is given as
$ = mc\Delta {T_C} = 100 \times 4.2 \times 10 = 4200J$
At equilibrium, heat lost by hot water will be equal to heat gained by cold water. Therefore, we have
$\begin{align}
& \Rightarrow 40 \times 4.2 \times (60 - x) = 4200 \\
& \Rightarrow (60 - x) = \dfrac{{4200}}{{4.2 \times 40}} = 25 \\
& \Rightarrow x = 60 - 25 = 35^\circ C \\
\end{align}$
This is the final temperature of the cold water. Therefore, we get its initial temperature in the following way:
Initial temperature $ = 35 - \Delta {T_C} = 35 - 10 = 25^\circ C$

Note:
 The final temperature of the mixture will be the same because when two liquids of different temperatures are mixed with each other, the colder one gains the heat energy while the hotter one loses the heat energy. At equilibrium, the temperatures of two liquids become equal.