Question

40 g of water at 40°C is added to 10 g of water at 80°C. The final temperature of the mixture is 40°C

• A. 48°C
• B. 40°C
• C. 120°C
• D. 64°C

Answer 1

The Correct Option is A

Let $m_1$ be the mass of water at $T_1$, and $m_2$ be the mass of water at $T_2$. 

We are given that $m_1 = 40$ g, $T_1 = 40^\circ C$, $m_2 = 10$ g, and $T_2 = 80^\circ C$. Let $T_f$ be the final temperature of the mixture. 

We can use the principle of calorimetry, which states that the heat lost by the hotter water is equal to the heat gained by the colder water. 

Let $c$ be the specific heat of water. 

Heat gained by the colder water is $Q_1 = m_1 c (T_f - T_1) = 40c(T_f - 40)$ 

Heat lost by the hotter water is $Q_2 = m_2 c (T_2 - T_f) = 10c(80 - T_f)$ 

According to the principle of calorimetry, $Q_1 = Q_2$, so $40c(T_f - 40) = 10c(80 - T_f)$ 

Since $c$ is a common factor, we can divide both sides by $10c$: $4(T_f - 40) = 80 - T_f$ $4T_f - 160 = 80 - T_f$ $5T_f = 240$ $T_f = \frac{240}{5} = 48$ 

So the final temperature of the mixture is $48^\circ C$.