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Calorimetry

Question

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 liters) are taken from each container and mixed (assume no loss of heat during the process)

The value of θ (in  0C  to the nearest integer) is -----------

Moderate
Solution

First Mixing : 1s θ160+2sθ260+0sθ360=0θ1+2θ2=180  . 1 Second Mixing:0s θ130+1sθ230+2sθ330=0θ2+2θ3=90     . 2 Third mixing: 2s θ160+0sθ260+1sθ360=02θ1+θ3=180   . 3 Solving equations 1 2 and 3 we get θ1=80, θ2=50, θ3=20 Fourth mixing: 1s θ1θ+1sθ2θ+1sθ3θ=0 80θ+50θ+20θ=0 θ=1503=50



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