A quantity of hot water at 100oC is added to 400g of water at 10oC until the temperature of the mixture is…

Analyze the question: Hot water at 100°C is mixed with cold water (400 g at 10°C). The final temperature is 60°C. We need the mass of the hot water.

Concept: When two bodies exchange heat with no loss to the surroundings:

Heat lost by hot water = Heat gained by cold water

Using Q = mcΔθ

Solution:

Let the mass of hot water be m (in g). Both are water, so c is the same on each side and cancels.

• Heat lost by hot water = m × c × (100 − 60) = m × c × 40
• Heat gained by cold water = 400 × c × (60 − 10) = 400 × c × 50

Equating:

m × 40 = 400 × 50

m = 20000 / 40 = 500 g

(Using SI units: 0.4 kg × 4200 × 50 = 84000 J gained, and m × 4200 × 40 = 84000 gives m = 0.5 kg = 500 g.)

This matches my working.

Answer: D. 500 g

Tip: If the same liquid is mixed, the specific heat capacity cancels out. Just write m₁Δθ₁ = m₂Δθ₂. Here the cold water rises by 50°C and the hot water drops by only 40°C, so the hot water mass must be larger than 400 g. That alone eliminates A, B and C.