Question

A 200 g sample of water at 80°C is mixed with 100 g of water at 20°C. Assuming no heat loss to the surroundings, what is the final temperature of the mixture?

• A. \( 50^\circ \text{C} \)
• B. \( 60^\circ \text{C} \)
• C. \( 55^\circ \text{C} \)
• D. \( 45^\circ \text{C} \)

Hint:

In mixing problems, use the principle of conservation of energy: heat lost by the hot body equals heat gained by the cold body. Remember to account for mass, specific heat, and temperature change.

Answer 1

The Correct Option is A

The heat energy transferred from the hotter water is equivalent to the heat energy absorbed by the cooler water. The relationship governing heat transfer is: \[ Q = mc\Delta T \] Where: - \( m \) represents the mass of the substance, - \( c \) denotes the specific heat capacity of water (\( 4.18 \, \text{J/g}^\circ\text{C} \)), - \( \Delta T \) indicates the variation in temperature. Let \( T_f \) be the final equilibrium temperature. For the hot water: \[ Q_{\text{hot}} = 200 \times 4.18 \times (80 - T_f) \] For the cold water: \[ Q_{\text{cold}} = 100 \times 4.18 \times (T_f - 20) \] Equating heat lost to heat gained: \[ 200 \times 4.18 \times (80 - T_f) = 100 \times 4.18 \times (T_f - 20) \] Simplifying the equation: \[ 200 \times (80 - T_f) = 100 \times (T_f - 20) \] \[ 16000 - 200T_f = 100T_f - 2000 \] \[ 16000 + 2000 = 300T_f \] \[ 18000 = 300T_f \] \[ T_f = \frac{18000}{300} = 60^\circ \text{C} \] Consequently, the mixture achieves a final temperature of \( 60^\circ \text{C} \).