Question

To a mass m1 = 100 g of water at θ1 = 10°C we add a mass m2 = 60 g of water at θ2 = 55°C. Calculate the final temperature of the mixture? Data: Specific heat capacity of water in the liquid state: 4.18 kJ.kg -1 .K

Answer 1

Given:

The mass of water is m1 = 100 g = 0.1 kg

The temperature of the water is

$\begin{gathered} \theta1=10^{\circ}\text{ C} \\ =10+273 \\ =283\text{ K} \end{gathered}$

The mass of water is m2 = 60g = 0.06 kg

The temperature of the water is

$\begin{gathered} \theta2=\text{ 55 }^{\circ}\text{C} \\ =55+273 \\ =328\text{ K} \end{gathered}$

To find the final temperature of the mixture.

Explanation:

The final temperature of the mixture can be calculated by the formula

$\begin{gathered} T=\frac{(m1\theta1+m2\theta2)}{(m1+m2)} \\ \text{ } \end{gathered}$

On substituting the values, the final temperature will be

$\begin{gathered} T=\frac{(0.1\times283)+(0.06\times328)}{(0.1+0.06)} \\ =293.625\text{ K} \\ =20.625\text{ }^{\circ}C \end{gathered}$

Thus, the final temperature of the mixture is 20.625 degrees Celsius.