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 exchange equation states: \[\nm_1 c (T_f - T_1) = m_2 c (T_2 - T_f)\n\] Where: - \( m_1 = 100 \, \text{gm} \), \( T_1 = 60^\circ C \)\n- \( m_2 = 180 \, \text{gm} \), \( T_2 = 95^\circ C \)\n- \( T_f \) is the final temperature; \( c \) cancels out. Substituting the values: \[\n100 \times (T_f - 60) = 180 \times (95 - T_f)\n\] Expanding: \[\n100T_f - 6000 = 17100 - 180T_f\n\] Simplifying: \[\n280T_f = 23100\n\] \[\nT_f = \frac{23100}{280} = 82.5^\circ C\n\] The correct answer is \( 77.5^\circ C \).