Ice of mass 10g at 5°C was completely converted to water at 0°C, calculate the quantity of heat used…

Analyze the question: Ice at some temperature is heated to 0°C, then melted completely at 0°C. The total heat is the heat to warm the ice plus the heat to melt it.

The question says the ice is at "5°C", but ice cannot exist at +5°C. The original WASSCE question almost certainly means -5°C. That is the only way the ice can be warmed to its melting point and then melted, so I'll solve it that way.

Concept:

• Heat to warm a substance: Q = mcΔθ
• Heat to change state: Q = mL

Solution:

Stage 1: Warm ice from -5°C to 0°C Q₁ = mcΔθ = 10 × 2.1 × 5 = 105 J

Stage 2: Melt ice at 0°C Q₂ = mL = 10 × 336 = 3360 J

Total heat: Q = Q₁ + Q₂ = 105 + 3360 = 3465 J

Check against the other options:

• A (105 J) is only the warming stage.
• B (3255 J) comes from subtracting 105 from 3360, which is wrong.
• D is too large to come from any correct step.

Exam tip: In "ice to water" questions, always list the stages: (1) heat the ice to 0°C, (2) melt it at 0°C, (3) heat the water if the final temperature is above 0°C. Add the heat from each stage. Remember that temperature stays constant during a change of state, so no mcΔθ term applies there.