Question

40 g of water at 60°C is added to. 60g of water at 30°C. The final temperature of their mixture is

• A. < 50°
• B. > 50°
• C. = 50°
• D. None of these

Answer 1

The Correct Option is A

Step 1: Use the principle of conservation of heat energy.

When two bodies of water at different temperatures are mixed, heat flows from the hotter body to the cooler body until thermal equilibrium is reached. The heat lost by the hotter water equals the heat gained by the cooler water:

\[ m_1 c \Delta T_1 = m_2 c \Delta T_2, \]

where:

• \( m_1 \) and \( m_2 \) are the masses of the two bodies of water,
• \( c \) is the specific heat capacity of water (same for both), and
• \( \Delta T_1 \) and \( \Delta T_2 \) are the changes in temperature for the two bodies.

 

Step 2: Set up the equation.

Let the final temperature of the mixture be \( T \). For the hotter water (\( 40 \, \text{g} \) at \( 60^\circ \text{C} \)):

\[ \Delta T_1 = 60 - T. \]

For the cooler water (\( 60 \, \text{g} \) at \( 30^\circ \text{C} \)):

\[ \Delta T_2 = T - 30. \]

Substitute into the heat balance equation:

\[ 40 (60 - T) = 60 (T - 30). \]

Step 3: Solve for \( T \).

Expand both sides:

\[ 2400 - 40T = 60T - 1800. \]

Rearrange terms:

\[ 2400 + 1800 = 60T + 40T \implies 4200 = 100T. \]

Solve for \( T \):

\[ T = \frac{4200}{100} = 42^\circ \text{C}. \]

Step 4: Compare with the options.

The final temperature (\( 42^\circ \text{C} \)) is less than \( 50^\circ \text{C} \).

Final Answer: The final temperature is \( \mathbf{< 50^\circ \text{C}} \), which corresponds to option \( \mathbf{(1)} \).