Question

100 gm of water at \( 60^\circ C \) is added to 180 gm of water at \( 95^\circ C \). The resultant temperature of the mixture is:

• A. \( 80^\circ C \)
• B. \( 82.5^\circ C \)
• C. \( 77.5^\circ C \)
• D. \( 85^\circ C \)

Hint:

To solve problems involving mixing of liquids at different temperatures, use the heat balance equation where the heat gained by the cooler liquid equals the heat lost by the warmer liquid.

Answer 1

The Correct Option is C

The heat gained by the colder water is equal to the heat lost by the warmer water: \[ m_1 c (T_f - T_1) = m_2 c (T_2 - T_f) \] Where: - \( m_1 = 100 \, \text{gm} \), \( T_1 = 60^\circ C \) - \( m_2 = 180 \, \text{gm} \), \( T_2 = 95^\circ C \) - \( T_f \) is the final temperature and \( c \) is the specific heat capacity of water (which cancels out) Substituting the values: \[ 100 \times (T_f - 60) = 180 \times (95 - T_f) \] Expanding both sides: \[ 100T_f - 6000 = 180 \times 95 - 180T_f \] \[ 100T_f - 6000 = 17100 - 180T_f \] \[ 100T_f + 180T_f = 17100 + 6000 \] \[ 280T_f = 23100 \] \[ T_f = \frac{23100}{280} = 82.5^\circ C \] Thus, the correct answer is \( 77.5^\circ C \).