Three containers C₁, C₂ and C₃ have water at different temperatures. The table below shows the final temperature T when different amounts of water (given in litres) are taken from each containers and mixed (assume no loss of heat during the process)
The value of θ (in °C to the nearest integer) is ..........

Solution

Physics sir · 4 steps

Same liquid, no heat lost: each mix is a weighted average, so add all three equations at once.

1. Weighted average. Let the containers be at θ₁, θ₂, θ₃. Every sample is water, so the specific heat cancels and the final temperature is the volume-weighted average of the starting temperatures.

   θ₁ + 2θ₂ = 3 × 60 = 180 ; θ₂ + 2θ₃ = 3 × 30 = 90 ; 2θ₁ + θ₃ = 3 × 60 = 180

2. Add them up. Adding the three equations gives 3(θ₁ + θ₂ + θ₃) = 450, so θ₁ + θ₂ + θ₃ = 150 °C. You do not need the individual values, but they are 80, 50 and 20 °C.
3. Equal mix. With 1 litre from each container, θ = (θ₁ + θ₂ + θ₃)/3 = 150/3.
4. So. θ = 50 °C.

The common slip: 150. Stopping at θ₁ + θ₂ + θ₃ = 150 gives 150. The equal mix is the average of the three, so divide by 3.

The idea to keep. When samples of the same liquid are mixed with no heat loss, the final temperature is the mass-weighted average. If the equations look symmetric, try adding them before solving each one.

Answer from the NTA final answer key. Solution by PrepWiser.