JAMB Physics 2018 Paper

Concept:Heat lost by the hot water equals heat gained by the cold water, assuming no heat is lost to the surroundings.Explanation:Let the final temperature of the mixture be $T$.Hot water: mass $= 150\,g = 0.15\,kg$, initial temperature $= 60^\circ C$.Heat lost by hot water: $Q_h = 0.15 c (60 - T)$.Cold water: mass $= 300\,g = 0.3\,kg$, initial temperature $= 20^\circ C$.Heat gained by cold water: $Q_c = 0.3 c (T - 20)$.Since $Q_h = Q_c$, we have:$0.15 c (60 - T) = 0.3 c (T - 20)$.The specific heat capacity $c$ cancels out.Expanding gives $9 - 0.15T = 0.3T - 6$.Rearranging: $0.3T + 0.15T = 9 + 6$.So $0.45T = 15$.Therefore, $T = \frac{15}{0.45} = 33.33^\circ C$.Answer:The temperature of the mixture is $33.33^\circ C$, approximately $33^\circ C$.Correct option: A. $33^\circ C$.